Exponential Functions and Equations

69 soru

Soru 1Soru

If 5x+25x=12055^{x+2} - 5^x = 120\sqrt{5}, what is the value of xx?

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Cevap: 1.5

Cevap

1.5 (or 3/2)
By factoring out the common term 5x5^x from the left side of the equation, we get 5x(521)=12055^x(5^2 - 1) = 120\sqrt{5}, which simplifies to 245x=120524 \cdot 5^x = 120\sqrt{5}. Dividing both sides by 24 isolates the exponential term: 5x=555^x = 5\sqrt{5}. Since 555\sqrt{5} can be written as 5150.5=51.55^1 \cdot 5^{0.5} = 5^{1.5}, we have 5x=51.55^x = 5^{1.5}. Equating the exponents gives x=1.5x = 1.5 (or 3/23/2).

Adım Adım Çözüm

1
Factor out 5x5^x from the left side of the equation.
5x(521)=12055^x(5^2 - 1) = 120\sqrt{5}
To apply exponent rules to rewrite 5x+25^{x+2} as 5x525^x \cdot 5^2 and then factor out the common term 5x5^x.
2
Simplify the constant term inside the parentheses.
245x=120524 \cdot 5^x = 120\sqrt{5}
Evaluating 521=251=245^2 - 1 = 25 - 1 = 24 simplifies the coefficient of the exponential expression.
3
Divide both sides of the equation by 24.
5x=555^x = 5\sqrt{5}
To isolate the exponential term 5x5^x on one side of the equation.
4
Express the right side as a single power of 5.
5x=51.55^x = 5^{1.5}
Using exponent rules where 55=5150.5=51.55\sqrt{5} = 5^1 \cdot 5^{0.5} = 5^{1.5} so that both sides have the same base.
5
Equate the exponents of the common base 5.
x=1.5x = 1.5
Since the bases on both sides of the equation are equal and positive, their exponents must be equal.

Anahtar Kavram

Solving exponential equations by factoring and rewriting terms using a common base.
Tahmini Süre:1m 30s
Soru 2Soru

If 27x=9x+127^x = 9^{x + 1}, what is the value of xx?

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Cevap: 2

Cevap

The value of xx is 22.
By rewriting 2727 as 333^3 and 99 as 323^2, the equation becomes (33)x=(32)x+1(3^3)^x = (3^2)^{x+1}. Applying the power rule of exponents, this simplifies to 33x=32x+23^{3x} = 3^{2x+2}. Since the bases are now the same, the exponents can be set equal to each other: 3x=2x+23x = 2x + 2. Solving for xx gives x=2x = 2.

Adım Adım Çözüm

1
Rewrite 2727 and 99 as powers of 33.
(33)x=(32)x+1(3^3)^x = (3^2)^{x + 1}
To solve exponential equations with different bases, it is helpful to express them using a common base.
2
Apply the power of a power rule (am)n=amn(a^m)^n = a^{mn} to simplify the exponents.
33x=32x+23^{3x} = 3^{2x + 2}
This simplifies the exponential expressions on both sides of the equation.
3
Set the exponents equal to each other.
3x=2x+23x = 2x + 2
Since the bases are equal (3=33 = 3), the exponents must also be equal for the equation to hold.
4
Solve the linear equation for xx.
x=2x = 2
Subtracting 2x2x from both sides isolates the variable xx.

Anahtar Kavram

Solving exponential equations by expressing both sides with a common base and equating exponents.
Tahmini Süre:45s
Soru 3Soru

If 4x+28x1=16x+14^{x+2} \cdot 8^{x-1} = 16^{x+1}, what is the value of xx?

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Cevap: 3

Cevap

3
To solve the equation 4x+28x1=16x+14^{x+2} \cdot 8^{x-1} = 16^{x+1}, rewrite all bases in terms of base 2: (22)x+2(23)x1=(24)x+1(2^2)^{x+2} \cdot (2^3)^{x-1} = (2^4)^{x+1}. Simplifying using the power rule yields 22x+423x3=24x+42^{2x+4} \cdot 2^{3x-3} = 2^{4x+4}. Applying the product rule on the left side gives 2(2x+4)+(3x3)=25x+12^{(2x+4)+(3x-3)} = 2^{5x+1}. Equating the exponents gives 5x+1=4x+45x + 1 = 4x + 4. Solving for xx results in x=3x = 3.

Adım Adım Çözüm

1
Express the bases 4, 8, and 16 as powers of 2.
4x+2=(22)x+2=22x+44^{x+2} = (2^2)^{x+2} = 2^{2x+4}, 8x1=(23)x1=23x38^{x-1} = (2^3)^{x-1} = 2^{3x-3}, and 16x+1=(24)x+1=24x+416^{x+1} = (2^4)^{x+1} = 2^{4x+4}
Writing all parts of the equation with a common base allows the exponents to be compared directly.
2
Combine the terms on the left side by adding their exponents.
22x+423x3=2(2x+4)+(3x3)=25x+12^{2x+4} \cdot 2^{3x-3} = 2^{(2x+4) + (3x-3)} = 2^{5x+1}
According to the product rule of exponents, bmbn=bm+nb^m \cdot b^n = b^{m+n} when the bases are the same.
3
Equate the exponents from both sides of the equation.
5x+1=4x+45x + 1 = 4x + 4
If two exponential expressions with the same positive base (other than 1) are equal, their exponents must be equal.
4
Solve the linear equation for xx.
x=3x = 3
Subtract 4x4x and 11 from both sides to isolate the variable xx.

Anahtar Kavram

Solving exponential equations by finding a common base and applying exponent laws.
Soru 4Soru

A sample of a radioactive isotope decays such that its mass, in grams, is modeled by the function M(t)=Adt12M(t) = A \cdot d^{\frac{t}{12}}, where tt is the time in hours since the measurement began, and AA and dd are positive constants. The mass of the sample decreases by 75%75\% every 2424 hours. If the mass of the sample is 1515 grams when t=36t = 36, what was the initial mass, in grams, of the sample?

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Cevap: 120

Cevap

The initial mass of the sample was 120 grams.
The correct answer is 120. A decrease of 75%75\% over 24 hours means the mass at t+24t+24 is 0.250.25 times the mass at tt. According to the function, M(t+24)=Ad(t+24)/12=Adt/12d2=M(t)d2M(t+24) = A \cdot d^{(t+24)/12} = A \cdot d^{t/12} \cdot d^2 = M(t) \cdot d^2. Equating the two yields d2=0.25d^2 = 0.25, so d=0.5d = 0.5. Substituting t=36t = 36 and M(36)=15M(36) = 15 gives 15=A(0.5)36/12=A(0.5)3=0.125A15 = A \cdot (0.5)^{36/12} = A \cdot (0.5)^3 = 0.125A. Solving for AA gives A=120A = 120.

Adım Adım Çözüm

1
Relate the 24-hour decay rate to the exponent in the function to set up an equation for dd.
d2=0.25d^2 = 0.25
Every 24 hours (tt increases by 24), the mass decreases by 75%75\%, so it becomes 25%25\% (0.250.25) of its previous value. The exponent increases by 24/12=224/12 = 2, multiplying the mass by d2d^2.
2
Solve for the decay base dd.
d=0.5d = 0.5
Since dd is a positive constant, taking the square root of 0.250.25 gives 0.50.5.
3
Substitute the given mass at t=36t = 36 into the model to solve for the initial mass AA.
A=120A = 120
Plugging t=36t = 36 and M(36)=15M(36) = 15 into M(t)=A(0.5)t/12M(t) = A \cdot (0.5)^{t/12} gives 15=A(0.5)315 = A \cdot (0.5)^3, which simplifies to 15=0.125A15 = 0.125A.

Anahtar Kavram

Interpreting and solving exponential decay functions with fractional exponents

Alternatif Yöntem

Instead of solving for dd first, recognize that 3636 hours is exactly 1.51.5 intervals of 2424 hours. Since the mass is multiplied by 0.250.25 (or 14\frac{1}{4}) every 2424 hours, after 3636 hours it will be multiplied by (14)1.5=(14)3/2=18(\frac{1}{4})^{1.5} = (\frac{1}{4})^{3/2} = \frac{1}{8} of its initial value. Therefore, 15=A18    A=12015 = A \cdot \frac{1}{8} \implies A = 120.
Tahmini Süre:3m 0s
Soru 5Soru

The amount of a radioactive isotope remaining after tt days is modeled by the function A(t)=A0(0.64)t2A(t) = A_0(0.64)^{\frac{t}{2}}, where A0A_0 is the initial amount of the isotope. If the function is rewritten in the form A(t)=A0(1r)tA(t) = A_0(1 - r)^t, where rr is the daily decay rate, what is the value of rr?

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Cevap: 0.20

Cevap

The daily decay rate rr is 0.200.20.
The correct value of rr is 0.200.20. By rewriting the given expression using exponent rules, we obtain (0.64)t/2=((0.64)1/2)t=(0.8)t(0.64)^{t/2} = ((0.64)^{1/2})^t = (0.8)^t. Setting the base equal to the target decay rate form gives 1r=0.81 - r = 0.8, which simplifies to r=0.20r = 0.20.

Adım Adım Çözüm

1
Apply the power of a power exponent rule, (xa)b=xab(x^a)^b = x^{ab}, to rewrite the exponential term.
A(t)=A0(0.641/2)tA(t) = A_0\left(0.64^{1/2}\right)^t
This isolates the variable tt as the exponent, allowing us to find the daily decay factor.
2
Evaluate the base 0.641/20.64^{1/2}.
0.64=0.8\sqrt{0.64} = 0.8, so the expression becomes A(t)=A0(0.8)tA(t) = A_0(0.8)^t.
Calculating the square root of 0.640.64 gives the daily decay factor of 0.80.8.
3
Equate the daily decay factor 0.80.8 to the target form 1r1 - r and solve for rr.
1r=0.8    r=0.201 - r = 0.8 \implies r = 0.20
Solving the equation gives the value of the daily decay rate.

Anahtar Kavram

Rewriting exponential equations by manipulating bases and exponents

Alternatif Yöntem

We can solve for rr by choosing a specific value for tt, such as t=2t = 2. After 2 days, the remaining amount is A(2)=A0(0.64)1=0.64A0A(2) = A_0(0.64)^1 = 0.64A_0. Using the target equation form, the remaining amount after 2 days is A0(1r)2A_0(1 - r)^2. Setting the two expressions equal to each other gives A0(1r)2=0.64A0A_0(1 - r)^2 = 0.64A_0. Dividing both sides by A0A_0 gives (1r)2=0.64(1 - r)^2 = 0.64. Taking the square root of both sides gives 1r=0.81 - r = 0.8, which yields r=0.20r = 0.20.
Tahmini Süre:1m 30s
Soru 6Soru

If 42n3=(18)n54^{2n-3} = \left(\frac{1}{8}\right)^{n-5}, what is the value of nn?

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Cevap: 3

Cevap

3
Rewriting the bases 4 and 18\frac{1}{8} as powers of 2 gives (22)2n3=(23)n5(2^2)^{2n-3} = (2^{-3})^{n-5}. Applying the exponent power rule simplifies this to 24n6=23n+152^{4n-6} = 2^{-3n+15}. Since the bases are equal, the exponents must be equal: 4n6=3n+154n-6 = -3n+15. Solving this linear equation by adding 3n3n and 6 to both sides gives 7n=217n = 21, which yields n=3n = 3.

Adım Adım Çözüm

1
Express the bases 4 and 18\frac{1}{8} as powers of the prime base 2.
4=224 = 2^2 and 18=23\frac{1}{8} = 2^{-3}
Rewriting the bases with a common base of 2 allows for the application of exponent rules to solve the equation.
2
Substitute these bases back into the original equation.
(22)2n3=(23)n5(2^2)^{2n-3} = (2^{-3})^{n-5}
This sets up the equation to simplify the exponent terms.
3
Apply the power of a power rule, (ab)c=abc(a^b)^c = a^{bc}, by multiplying the exponents on both sides.
22(2n3)=23(n5)    24n6=23n+152^{2(2n-3)} = 2^{-3(n-5)} \implies 2^{4n-6} = 2^{-3n+15}
Multiplying the exponents simplifies the expressions to a single base raised to a single exponent on each side.
4
Equate the exponents since the bases are now equal, and solve for nn.
4n6=3n+15    7n=21    n=34n-6 = -3n+15 \implies 7n = 21 \implies n = 3
If bx=byb^x = b^y where b>0b > 0 and b1b \neq 1, then x=yx = y.

Anahtar Kavram

Solving exponential equations by expressing bases in terms of a common base and applying exponent rules.
Soru 7Soru

If 3x2=813^{x - 2} = 81, what is the value of xx?

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Cevap: 6

Cevap

6
To solve the equation 3x2=813^{x - 2} = 81, we first express the number 81 as a power of 3, which is 343^4. This gives us 3x2=343^{x - 2} = 3^4. Since the bases are equal, we can set their exponents equal to each other, resulting in the equation x2=4x - 2 = 4. Solving for xx by adding 2 to both sides gives the correct value of 6.

Adım Adım Çözüm

1
Express both sides of the equation with a common base of 3.
3x2=343^{x - 2} = 3^4
Since 81 is equal to 3×3×3×33 \times 3 \times 3 \times 3, it can be written as 343^4.
2
Set the exponents equal to each other because the bases are now the same.
x2=4x - 2 = 4
If by=bzb^y = b^z for a positive base b1b \neq 1, then y=zy = z.
3
Solve the linear equation for xx by adding 2 to both sides of the equation.
x=6x = 6
Adding 2 to both sides isolates the variable xx.

Anahtar Kavram

Solving exponential equations by expressing both sides with a common base and equating their exponents.
Soru 8Soru

A population of bacteria doubles every 3 hours. If the initial population of the bacteria is 500, what is the population of the bacteria after 9 hours?

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Cevap: 4000

Cevap

The population of the bacteria after 9 hours is 4,000.
The final population is calculated using the formula P(t)=P0×2t/dP(t) = P_0 \times 2^{t/d}, where P0P_0 is the initial population of 500, dd is the doubling period of 3 hours, and tt is the total time of 9 hours. Evaluating this gives P(9)=500×29/3=500×23=500×8=4000P(9) = 500 \times 2^{9/3} = 500 \times 2^3 = 500 \times 8 = 4000.

Adım Adım Çözüm

1
Identify the initial population (P0P_0), doubling time (dd), and total time (tt).
P0=500P_0 = 500, d=3d = 3, and t=9t = 9.
These parameters are required to set up the exponential growth model.
2
Calculate the number of doubling periods.
The number of doubling periods is 93=3\frac{9}{3} = 3.
The population doubles once for every 3-hour interval.
3
Calculate the final population using the exponential growth formula.
500×23=500×8=4000500 \times 2^3 = 500 \times 8 = 4000.
Applying the 3 doubling cycles to the initial population of 500 yields the final population.

Anahtar Kavram

Exponential Growth Model
Tahmini Süre:1m 0s
Soru 9Soru

A bank account is opened with an initial deposit of 800.Theaccountbalanceincreasesby800. The account balance increases by 5\%eachyear.Ifnoothertransactionsaremade,whichofthefollowingfunctionsbestmodelstheaccountbalance, each year. If no other transactions are made, which of the following functions best models the account balance, B(t),indollars,after, in dollars, after t$ years?

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Cevap: B(t)=800(1.05)tB(t) = 800(1.05)^t

Cevap

The function B(t)=800(1.05)tB(t) = 800(1.05)^t best models the account balance.
The initial deposit of 800representstheinitialvalueoftheexponentialfunctionwhen800 represents the initial value of the exponential function when t = 0 .Sincetheaccountbalanceincreasesby. Since the account balance increases by 5\%eachyear,thevalueismultipliedbyagrowthfactorof each year, the value is multiplied by a growth factor of 1 + 0.05 = 1.05 eachyear.Thus,theexponentialfunctionthatmodelsthebalanceafter each year. Thus, the exponential function that models the balance after t yearsis years is B(t) = 800(1.05)^t$.

Adım Adım Çözüm

1
Identify the initial value of the exponential growth function.
The initial deposit is 800,sotheinitialvalueat800, so the initial value at t = 0 is is 800$.
An exponential model is written in the form B(t)=a(b)tB(t) = a(b)^t, where aa represents the initial value.
2
Determine the growth factor based on the annual percentage increase.
The growth rate is r=5%=0.05r = 5\% = 0.05. The growth factor bb is 1+r=1+0.05=1.051 + r = 1 + 0.05 = 1.05.
To find the growth factor for an increasing quantity, add the growth rate as a decimal to 1.
3
Write the final exponential function by substituting the initial value and growth factor.
The function is B(t)=800(1.05)tB(t) = 800(1.05)^t.
Substitute a=800a = 800 and b=1.05b = 1.05 into the standard exponential form B(t)=a(b)tB(t) = a(b)^t.

Anahtar Kavram

Writing and interpreting exponential growth functions from a real-world context.
Soru 10Soru

A scientist studying a sample of a radioactive isotope determines that its mass decays exponentially. The mass of the isotope, in grams, is modeled by the function M(t)=abtM(t) = a \cdot b^t, where tt is the time, in days, since the study began, and aa and bb are constants. The table below shows the mass of the isotope for selected values of tt.

tt (days)M(t)M(t) (grams)
0128
196
272

What is the value of bb?

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Cevap: 0.75

Cevap

0.75
To find the constant bb in the exponential decay function M(t)=abtM(t) = a \cdot b^t, we can use the given table values. For t=0t = 0, M(0)=ab0=a=128M(0) = a \cdot b^0 = a = 128. For t=1t = 1, M(1)=ab1=ab=96M(1) = a \cdot b^1 = a \cdot b = 96. Substituting a=128a = 128 into the second equation gives 128b=96128b = 96. Dividing both sides by 128128 yields b=96128=0.75b = \frac{96}{128} = 0.75 (or 34\frac{3}{4}). We can verify this with t=2t = 2: M(2)=128(0.75)2=1280.5625=72M(2) = 128 \cdot (0.75)^2 = 128 \cdot 0.5625 = 72, which matches the table.

Adım Adım Çözüm

1
Determine the initial value aa using the data point for t=0t = 0.
a=128a = 128
Substituting t=0t = 0 into M(t)=abtM(t) = a \cdot b^t gives M(0)=ab0=aM(0) = a \cdot b^0 = a. Since the table shows M(0)=128M(0) = 128, aa must equal 128128.
2
Use the data point for t=1t = 1 to write an equation for bb.
128b=96128b = 96
Substituting t=1t = 1 and a=128a = 128 into the function gives M(1)=128b1=128bM(1) = 128 \cdot b^1 = 128b. The table shows M(1)=96M(1) = 96.
3
Solve the equation to find bb.
b=0.75b = 0.75
Dividing both sides of 128b=96128b = 96 by 128128 yields b=96128b = \frac{96}{128}, which simplifies to 0.750.75 (or 34\frac{3}{4}).

Anahtar Kavram

Finding the decay factor (base) of an exponential function from a table of values.
Soru 11Soru

If 92x+1=27x39^{2x + 1} = 27^{x - 3}, what is the value of xx?

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Cevap: 11-11

Cevap

11-11
To solve 92x+1=27x39^{2x + 1} = 27^{x - 3}, express 9 and 27 with a common base of 3: (32)2x+1=(33)x3(3^2)^{2x + 1} = (3^3)^{x - 3}. Apply the power of a power rule to get 32(2x+1)=33(x3)3^{2(2x + 1)} = 3^{3(x - 3)}, which simplifies to 34x+2=33x93^{4x + 2} = 3^{3x - 9}. Equating the exponents gives 4x+2=3x94x + 2 = 3x - 9. Solving for xx by subtracting 3x3x and 22 from both sides yields x=11x = -11. Therefore, the value of xx is 11-11.

Adım Adım Çözüm

1
Rewrite both sides of the equation with a common base.
(32)2x+1=(33)x3(3^2)^{2x+1} = (3^3)^{x-3}
Since 9 and 27 are both powers of 3 (9=329 = 3^2 and 27=3327 = 3^3), expressing them with the same base allows the application of exponent rules.
2
Apply the power of a power rule (am)n=amn(a^m)^n = a^{mn} to simplify the exponents.
32(2x+1)=33(x3)3^{2(2x+1)} = 3^{3(x-3)}, which simplifies to 34x+2=33x93^{4x+2} = 3^{3x-9}
Multiplying the inner exponent by the outer exponent simplifies the expression on both sides.
3
Set the exponents equal to each other.
4x+2=3x94x + 2 = 3x - 9
Since the bases are equal and positive, the exponential expressions are equal if and only if their exponents are equal.
4
Solve the linear equation for xx.
x=11x = -11
Subtracting 3x3x and 22 from both sides isolates the variable xx.

Anahtar Kavram

Solving exponential equations by converting to a common base and equating exponents.
Soru 12Soru

A wildlife biologist models the population of a certain bird species in a nature reserve using the function P(t)=120(1.05)tP(t) = 120(1.05)^t, where P(t)P(t) represents the number of birds tt years after the study began. What does the number 120120 represent in this model?

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Cevap: The initial number of birds in the reserve when the study began

Cevap

The initial number of birds in the reserve when the study began
In the exponential function model P(t)=abtP(t) = a \cdot b^t, the coefficient aa is the value of the function at t=0t = 0. Evaluating P(0)P(0) yields 120(1.05)0=120120(1.05)^0 = 120, which corresponds to the initial number of birds in the reserve when the study began.

Adım Adım Çözüm

1
Identify the standard form of the exponential growth function
P(t)=abtP(t) = a \cdot b^t, where aa represents the initial value and bb represents the growth factor.
This helps map the given constants to their mathematical meanings.
2
Substitute t=0t = 0 to find the starting population
P(0)=120(1.05)0=1201=120P(0) = 120(1.05)^0 = 120 \cdot 1 = 120.
Finding the value at t=0t = 0 defines the initial state of the model.
3
Interpret the initial value in the context of the problem
The number 120120 represents the number of birds present at the start of the study.
This links the mathematical result to the real-world scenario.

Anahtar Kavram

Interpreting components of an exponential function in context
Soru 13Soru

If 8x+2=(14)13x8^{x+2} = \left(\frac{1}{4}\right)^{1-3x}, what is the value of xx?

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Cevap: 83\frac{8}{3}

Cevap

83\frac{8}{3}
To solve the equation 8x+2=(14)13x8^{x+2} = \left(\frac{1}{4}\right)^{1-3x}, both bases can be written as powers of 22 because 8=238 = 2^3 and 14=22\frac{1}{4} = 2^{-2}. Substituting these values into the equation yields (23)x+2=(22)13x(2^3)^{x+2} = (2^{-2})^{1-3x}. Applying the power rule of exponents, (am)n=amn(a^m)^n = a^{mn}, we multiply the exponents to get 23x+6=22+6x2^{3x+6} = 2^{-2+6x}. Since the bases are equal, we can set the exponents equal to each other: 3x+6=2+6x3x + 6 = -2 + 6x. Solving this linear equation by subtracting 3x3x from both sides and adding 22 to both sides gives 8=3x8 = 3x, which simplifies to x=83x = \frac{8}{3}. Thus, the option with the value 83\frac{8}{3} is correct.

Adım Adım Çözüm

1
Express both bases in the equation, 88 and 14\frac{1}{4}, as powers of 22.
The base 88 is written as 232^3 and the base 14\frac{1}{4} is written as 222^{-2}, yielding the equation (23)x+2=(22)13x(2^3)^{x+2} = (2^{-2})^{1-3x}.
Expressing exponential terms with a common base is necessary to equate and solve their exponents.
2
Apply the power of a power exponent rule, (am)n=amn(a^m)^n = a^{mn}, to simplify both sides of the equation.
The equation becomes 23(x+2)=22(13x)2^{3(x+2)} = 2^{-2(1-3x)}, which simplifies to 23x+6=22+6x2^{3x+6} = 2^{-2+6x}.
This simplifies each side to a single base with a single exponent.
3
Since the bases are equal, set the exponents equal to each other and solve the resulting linear equation for xx.
3x+6=2+6x    8=3x    x=833x + 6 = -2 + 6x \implies 8 = 3x \implies x = \frac{8}{3}.
Two exponential expressions with the same positive base are equal if and only if their exponents are equal.

Anahtar Kavram

Solving exponential equations by expressing terms with a common base and applying exponent rules.
Soru 14Soru

If 25a=125b25^a = 125^b, where aa and bb are positive constants, what is the value of ab\frac{a}{b}?

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Cevap: 1.5

Cevap

1.5
To find the value of ab\frac{a}{b}, we can rewrite the bases of the equation 25a=125b25^a = 125^b using a common base of 5: (52)a=(53)b(5^2)^a = (5^3)^b. Applying the exponent power rule gives 52a=53b5^{2a} = 5^{3b}. Since the bases are identical, their exponents must be equal, so 2a=3b2a = 3b. Dividing both sides of this equation by 2b2b yields the ratio ab=32\frac{a}{b} = \frac{3}{2}, which is 1.51.5.

Adım Adım Çözüm

1
Express the bases as powers of 5
(52)a=(53)b(5^2)^a = (5^3)^b
To solve the equation, express both sides with a common base of 5 since 25=5225 = 5^2 and 125=53125 = 5^3.
2
Apply the power rule of exponents
52a=53b5^{2a} = 5^{3b}
According to the exponent rules, (xm)n=xmn(x^m)^n = x^{mn}.
3
Set the exponents equal to each other
2a=3b2a = 3b
Because the bases on both sides of the equation are equal and positive, their exponents must also be equal.
4
Solve for the ratio
ab=1.5\frac{a}{b} = 1.5
Divide both sides of the equation 2a=3b2a = 3b by 2b2b to isolate the ratio ab\frac{a}{b}.

Anahtar Kavram

Solving exponential equations by expressing bases in terms of a common base and equating exponents.
Soru 15Soru

The function ff is defined by f(x)=a3xf(x) = a \cdot 3^x, where aa is a constant. If f(2)=45f(2) = 45, what is the value of f(1)f(1)?

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Cevap: 15

Cevap

15
To find the value of f(1)f(1), we first determine the value of the constant aa. We are given that f(2)=45f(2) = 45, so substituting x=2x = 2 into the function definition f(x)=a3xf(x) = a \cdot 3^x gives 45=a3245 = a \cdot 3^2. Simplifying 323^2 to 99 yields 45=9a45 = 9a, which means a=5a = 5. Now that we know a=5a = 5, we can write the function as f(x)=53xf(x) = 5 \cdot 3^x. To find f(1)f(1), we substitute x=1x = 1 into this equation, yielding f(1)=531=15f(1) = 5 \cdot 3^1 = 15.

Adım Adım Çözüm

1
Substitute the given point (2,45)(2, 45) into the function equation to solve for aa.
a=5a = 5
Since f(2)=45f(2) = 45, we have 45=a32=9a45 = a \cdot 3^2 = 9a, which gives a=5a = 5.
2
Evaluate the function at x=1x = 1 using the value of a=5a = 5.
f(1)=15f(1) = 15
Substituting a=5a = 5 and x=1x = 1 into f(x)=a3xf(x) = a \cdot 3^x gives f(1)=531=15f(1) = 5 \cdot 3^1 = 15.

Anahtar Kavram

Evaluating and solving exponential functions given initial conditions or points.
Soru 16Soru

If 3x+1=k3^{x+1} = k, where k>0k > 0, which of the following is equivalent to 27x127^{x-1}?

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Cevap: k3729\frac{k^3}{729}

Cevap

Theexpressionk3729The expression \frac{k^3}{729}
To express 27x127^{x-1} in terms of kk, we can rewrite both the given equation and the target expression using base 33. The given equation 3x+1=k3^{x+1} = k can be adjusted to find the value of 3x13^{x-1} by dividing both sides by 323^2, which yields 3x1=k93^{x-1} = \frac{k}{9}. The target expression 27x127^{x-1} can be rewritten as (33)x1=(3x1)3(3^3)^{x-1} = (3^{x-1})^3. Substituting 3x1=k93^{x-1} = \frac{k}{9} into this expression gives (k9)3=k3729\left(\frac{k}{9}\right)^3 = \frac{k^3}{729}.

Adım Adım Çözüm

1
Express the given equation in terms of a simpler base 33 exponent.
3x1=k93^{x-1} = \frac{k}{9}
We start with the given equation 3x+1=k3^{x+1} = k. To relate this to the exponent x1x-1, we divide both sides of the equation by 32=93^2 = 9: 3x1=3x+132=k93^{x-1} = \frac{3^{x+1}}{3^2} = \frac{k}{9}.
2
Rewrite the target expression 27x127^{x-1} with base 33.
27x1=(3x1)327^{x-1} = (3^{x-1})^3
Since 27=3327 = 3^3, we can rewrite the target expression as 27x1=(33)x1=33(x1)=(3x1)327^{x-1} = (3^3)^{x-1} = 3^{3(x-1)} = (3^{x-1})^3.
3
Substitute the expression for 3x13^{x-1} from Step 1 into the expression from Step 2.
k3729\frac{k^3}{729}
Substituting 3x1=k93^{x-1} = \frac{k}{9} into (3x1)3(3^{x-1})^3 gives (k9)3=k393=k3729\left(\frac{k}{9}\right)^3 = \frac{k^3}{9^3} = \frac{k^3}{729}.

Anahtar Kavram

Manipulating exponential equations by expressing terms with a common base and applying exponent rules.
Soru 17Soru

If 22x+222x=122^{2x+2} - 2^{2x} = 12 for some real number xx, what is the value of 24x2^{4x}?

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Cevap: 16

Cevap

16
By applying the product rule of exponents, the expression 22x+22^{2x+2} can be rewritten as 22x222^{2x} \cdot 2^2, or 422x4 \cdot 2^{2x}. Substituting this into the given equation yields 422x22x=124 \cdot 2^{2x} - 2^{2x} = 12. Factoring out 22x2^{2x} gives 22x(41)=122^{2x}(4 - 1) = 12, which simplifies to 322x=123 \cdot 2^{2x} = 12. Dividing both sides of the equation by 3 results in 22x=42^{2x} = 4. Since 24x2^{4x} can be written as (22x)2(2^{2x})^2, substituting 4 for 22x2^{2x} gives 42=164^2 = 16. Alternatively, solving 22x=42^{2x} = 4 gives 2x=22x = 2, which means x=1x = 1. Substituting x=1x = 1 into 24x2^{4x} yields 24(1)=24=162^{4(1)} = 2^4 = 16.

Adım Adım Çözüm

1
Rewrite 22x+22^{2x+2} using exponent properties.
22x+2=22x22=422x2^{2x+2} = 2^{2x} \cdot 2^2 = 4 \cdot 2^{2x}
To express both exponential terms with the same base power, allowing them to be factored or combined.
2
Substitute this back into the equation and factor out the common term 22x2^{2x}.
22x(41)=12    322x=122^{2x}(4 - 1) = 12 \implies 3 \cdot 2^{2x} = 12
To isolate the exponential expression.
3
Solve for 22x2^{2x} by dividing both sides of the equation by 3.
22x=42^{2x} = 4
To find the value of the exponential term.
4
Express 24x2^{4x} in terms of 22x2^{2x} and evaluate.
24x=(22x)2=42=162^{4x} = (2^{2x})^2 = 4^2 = 16
To find the final requested value using the exponent rule (am)n=amn(a^m)^n = a^{mn}.

Anahtar Kavram

Factoring exponential equations and applying power of a power exponent rules
Soru 18Soru

If 3x+4=9x3^{x+4} = 9^x, what is the value of xx?

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Cevap: 4

Cevap

4
To solve the equation 3x+4=9x3^{x+4} = 9^x, we rewrite the base 9 as 323^2, which gives 3x+4=(32)x3^{x+4} = (3^2)^x. Applying the exponent rule (am)n=amn(a^m)^n = a^{mn}, we get 3x+4=32x3^{x+4} = 3^{2x}. Since the bases are now the same, we can set the exponents equal to each other: x+4=2xx+4 = 2x. Subtracting xx from both sides yields the solution 44.

Adım Adım Çözüm

1
Express both sides of the equation with a common base.
Since 9=329 = 3^2, the equation 3x+4=9x3^{x+4} = 9^x can be rewritten as 3x+4=(32)x3^{x+4} = (3^2)^x. Using the power of a power rule, this becomes 3x+4=32x3^{x+4} = 3^{2x}.
To solve an exponential equation, it is helpful to have the same base on both sides so that the exponents can be equated.
2
Set the exponents equal to each other.
x+4=2xx + 4 = 2x
Since the bases are equal and positive (and not equal to 1), their exponents must be equal.
3
Solve the linear equation for xx.
4=x4 = x (or x=4x = 4)
Subtract xx from both sides of the equation to isolate the variable.

Anahtar Kavram

Solving exponential equations by expressing terms with a common base and equating their exponents.
Tahmini Süre:45s
Soru 19Soru

If 4x1=84^{x - 1} = 8, what is the value of xx?

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Cevap: 2.5

Cevap

2.5 (or 5/2)
The correct answer is 2.5 (or 5/2). By expressing 4 as 222^2 and 8 as 232^3, the equation is rewritten as (22)x1=23(2^2)^{x-1} = 2^3. Applying the exponent power rule simplifies this to 22x2=232^{2x-2} = 2^3. Equating the exponents yields the linear equation 2x2=32x-2 = 3. Solving for xx gives 2x=52x = 5, which results in x=2.5x = 2.5 or 5/25/2.

Adım Adım Çözüm

1
Rewrite both sides of the equation with a common base of 2.
(22)x1=23(2^2)^{x-1} = 2^3
Expressing both bases as powers of 2 allows us to equate the exponents later.
2
Apply the power of a power exponent rule (am)n=amn(a^m)^n = a^{mn} to the left side.
22x2=232^{2x-2} = 2^3
Multiplying the exponent 2 by the exponent (x1)(x-1) simplifies the expression to 2(x1)=2x22(x-1) = 2x-2.
3
Equate the exponents and solve the linear equation for xx.
x=2.5x = 2.5
Since the bases are identical, their exponents must be equal, giving 2x2=32x-2 = 3, which simplifies to 2x=52x = 5.

Anahtar Kavram

Solving exponential equations by finding a common base.
Soru 20Soru

If 32x1=273^{2x - 1} = 27, what is the value of xx?

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Cevap: 2

Cevap

2
To solve the equation 32x1=273^{2x - 1} = 27, we first express 27 as a power of 3, which is 333^3. Since the bases are the same, we set the exponents equal to each other to get 2x1=32x - 1 = 3. Adding 1 to both sides gives 2x=42x = 4, and dividing both sides by 2 gives the value of xx as 2.

Adım Adım Çözüm

1
Express 27 with a base of 3
32x1=333^{2x - 1} = 3^3
To solve an exponential equation, we need to write both sides of the equation with a common base.
2
Set the exponents equal to each other
2x1=32x - 1 = 3
Since the bases are equal, their exponents must also be equal.
3
Solve the linear equation for xx
x=2x = 2
Add 1 to both sides to get 2x=42x = 4, and then divide both sides by 2 to isolate xx.

Anahtar Kavram

Solving exponential equations by finding a common base
Tahmini Süre:45s
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Exponential Functions and Equations Alıştırma Soruları — SAT | Examkin