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Logarithms and Change of Base

24 questions

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Question 21Question →

If log⁡2x+log⁡4x+log⁡16x=7\log_2 x + \log_4 x + \log_{16} x = 7, what is the value of xx?

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Answer: 1616

Answer

1616
Using the change of base property, log⁡4x=12log⁡2x\log_4 x = \frac{1}{2}\log_2 x and log⁡16x=14log⁡2x\log_{16} x = \frac{1}{4}\log_2 x. Combining like terms yields (1+12+14)log⁡2x=74log⁡2x\left(1 + \frac{1}{2} + \frac{1}{4}\right)\log_2 x = \frac{7}{4}\log_2 x. Setting 74log⁡2x=7\frac{7}{4}\log_2 x = 7 gives log⁡2x=4\log_2 x = 4, which leads to x=24=16x = 2^4 = 16.

Step-by-Step Solution

1
Apply the change of base formula to express all logarithmic terms in base 2.
log⁡4x=log⁡2xlog⁡24=12log⁡2x\log_4 x = \frac{\log_2 x}{\log_2 4} = \frac{1}{2}\log_2 x and log⁡16x=log⁡2xlog⁡216=14log⁡2x\log_{16} x = \frac{\log_2 x}{\log_2 16} = \frac{1}{4}\log_2 x.
Logarithms with different bases must be converted to a common base to combine them.
2
Substitute the transformed terms back into the equation and factor out log⁡2x\log_2 x.
\log_2 x + \frac{1}{2}\log_2 x + \frac{1}{4}\log_2 x = \left(1 + \frac{1}{2} + \frac{1}{4}\right)\log_2 x = \frac{7}{4}\log_2 x.
Factoring allows summing the coefficients of log⁡2x\log_2 x.
3
Solve for log⁡2x\log_2 x by equating the simplified expression to 7.
\frac{7}{4}\log_2 x = 7 \implies \log_2 x = 7 \times \frac{4}{7} = 4.
Isolating the logarithmic term gives its numerical value.
4
Convert the logarithmic equation to exponential form to solve for xx.
x = 2^4 = 16.
By definition, log⁡ba=c  ⟺  a=bc\log_b a = c \iff a = b^c.

Key Concept

Change of Base Formula for Logarithms
Estimated Time:1m 30s
Question 22Question →

If log⁡3x−2log⁡x27=1\log_3 x - 2\log_x 27 = 1, what is the sum of all possible real values of xx?

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Answer: 2449\frac{244}{9}

Answer

2449\frac{244}{9}
By applying the change of base rule log⁡x27=log⁡327log⁡3x=3log⁡3x\log_x 27 = \frac{\log_3 27}{\log_3 x} = \frac{3}{\log_3 x}, the given equation simplifies to log⁡3x−6log⁡3x=1\log_3 x - \frac{6}{\log_3 x} = 1. Setting u=log⁡3xu = \log_3 x yields u2−u−6=0u^2 - u - 6 = 0, which factors as (u−3)(u+2)=0(u - 3)(u + 2) = 0. Thus, u=3u = 3 or u=−2u = -2, giving solutions x=33=27x = 3^3 = 27 and x=3−2=19x = 3^{-2} = \frac{1}{9}. Adding these valid real solutions gives 27+19=244927 + \frac{1}{9} = \frac{244}{9}.

Step-by-Step Solution

1
Apply the change of base formula to log⁡x27\log_x 27.
log⁡x27=log⁡327log⁡3x=3log⁡3x\log_x 27 = \frac{\log_3 27}{\log_3 x} = \frac{3}{\log_3 x}
Converting all logarithmic terms to a common base (base 3) allows substitution.
2
Substitute log⁡x27\log_x 27 back into the original equation and let u=log⁡3xu = \log_3 x.
u−2(3u)=1  ⟹  u−6u=1u - 2\left(\frac{3}{u}\right) = 1 \implies u - \frac{6}{u} = 1
This transforms the logarithmic equation into an algebraic equation.
3
Clear the denominator by multiplying through by uu and rearrange into standard quadratic form.
u2−u−6=0u^2 - u - 6 = 0
Multiplying by uu (where u≠0u \neq 0) forms a standard quadratic equation.
4
Factor the quadratic equation to solve for uu.
(u−3)(u+2)=0  ⟹  u=3 or u=−2(u - 3)(u + 2) = 0 \implies u = 3 \text{ or } u = -2
Factoring determines the values of log⁡3x\log_3 x.
5
Convert back to xx using x=3ux = 3^u.
For u=3u = 3, x=33=27x = 3^3 = 27. For u=−2u = -2, x=3−2=19x = 3^{-2} = \frac{1}{9}. Both x=27x = 27 and x=19x = \frac{1}{9} are valid bases (x>0,x≠1x > 0, x \neq 1).
Exponentiation resolves the logarithmic variable.
6
Find the sum of all valid values of xx.
Sum=27+19=243+19=2449\text{Sum} = 27 + \frac{1}{9} = \frac{243 + 1}{9} = \frac{244}{9}
Combines the two real solutions into the final requested sum.

Key Concept

Change of base formula log⁡ab=log⁡cblog⁡ca\log_a b = \frac{\log_c b}{\log_c a} and solving equations reducible to quadratics.
Estimated Time:2m 0s
Question 23Question →

If log⁡2x+log⁡8x=4\log_2 x + \log_8 x = 4, what is the value of xx?

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Answer: 8

Answer

8
By applying the change of base formula, log⁡8x=log⁡2xlog⁡28=13log⁡2x\log_8 x = \frac{\log_2 x}{\log_2 8} = \frac{1}{3}\log_2 x. Rewriting the equation gives log⁡2x+13log⁡2x=43log⁡2x=4\log_2 x + \frac{1}{3}\log_2 x = \frac{4}{3}\log_2 x = 4. Solving for log⁡2x\log_2 x yields log⁡2x=3\log_2 x = 3, which in exponential form gives x=23=8x = 2^3 = 8.

Step-by-Step Solution

1
Apply the change of base formula to express log⁡8x\log_8 x in base 2
log⁡8x=log⁡2xlog⁡28=log⁡2x3\log_8 x = \frac{\log_2 x}{\log_2 8} = \frac{\log_2 x}{3}
Logarithms must be converted to a common base before combining terms.
2
Substitute log⁡8x=13log⁡2x\log_8 x = \frac{1}{3}\log_2 x into the original equation
log⁡2x+13log⁡2x=4  ⟹  43log⁡2x=4\log_2 x + \frac{1}{3}\log_2 x = 4 \implies \frac{4}{3}\log_2 x = 4
Combine like logarithmic terms.
3
Solve for log⁡2x\log_2 x
log⁡2x=4×34=3\log_2 x = 4 \times \frac{3}{4} = 3
Isolate the logarithmic expression by multiplying both sides by 34\frac{3}{4}.
4
Convert the logarithmic equation to its exponential form to solve for xx
x=23=8x = 2^3 = 8
Definition of logarithms states that log⁡ba=c  ⟺  bc=a\log_b a = c \iff b^c = a.

Key Concept

Logarithms and Change of Base
Question 24Question →

What is the value of 1log⁡2100+1log⁡5100\frac{1}{\log_2 100} + \frac{1}{\log_5 100}?

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Answer: 12\frac{1}{2}

Answer

12\frac{1}{2}
Using the change of base identity 1log⁡ab=log⁡ba\frac{1}{\log_a b} = \log_b a, the expression simplifies to log⁡1002+log⁡1005\log_{100} 2 + \log_{100} 5. By the product rule of logarithms, this equals log⁡100(2×5)=log⁡10010\log_{100}(2 \times 5) = \log_{100} 10. Since 10012=10100^{\frac{1}{2}} = 10, the value is 12\frac{1}{2}.

Step-by-Step Solution

1
Apply the reciprocal change of base rule 1log⁡ab=log⁡ba\frac{1}{\log_a b} = \log_b a.
1log⁡2100=log⁡1002\frac{1}{\log_2 100} = \log_{100} 2 and 1log⁡5100=log⁡1005\frac{1}{\log_5 100} = \log_{100} 5.
Converting to a common base of 100100 allows the use of logarithmic addition laws.
2
Combine the two logarithms using the product law log⁡cx+log⁡cy=log⁡c(xy)\log_c x + \log_c y = \log_c (xy).
\log_{100} 2 + \log_{100} 5 = \log_{100} (2 \times 5) = \log_{100} 10.
Adding logarithms with the same base is equivalent to taking the logarithm of the product of their arguments.
3
Evaluate log⁡10010\log_{100} 10.
Since 10012=10100^{\frac{1}{2}} = 10, log⁡10010=12\log_{100} 10 = \frac{1}{2}.
The logarithm asks what power base 100100 must be raised to in order to equal 1010.

Key Concept

Logarithms and Change of Base
Estimated Time:1m 30s
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