Exponents, Powers, and Square Roots Alıştırma Soruları — GRE General Test — Sayfa 2

If

x

is a positive integer such that

\frac{4^x + 4^x + 4^x + 4^x}{2^x + 2^x} = 512

what is the value of

x

?

Cevabı ve açıklamayı göster

Cevap: 8

Cevap

The value of

x

is 8.

Combining four terms of

4^x

yields

4 \cdot 4^x = 4^{x+1} = 2^{2x+2}

. Combining two terms of

2^x

yields

2 \cdot 2^x = 2^{x+1}

. Dividing the numerator by the denominator gives

2^{2x+2 - (x+1)} = 2^{x+1}

. Since

512 = 2^9

, setting

2^{x+1} = 2^9

gives

x + 1 = 9

, which leads directly to

x = 8

.

Adım Adım Çözüm

1

Simplify the numerator by combining identical added terms.

The numerator

4^x + 4^x + 4^x + 4^x

equals

4 \cdot 4^x

, which simplifies to

4^{x+1}

.

Adding four identical quantities is equivalent to multiplying that quantity by 4.

2

Simplify the denominator by combining identical added terms.

The denominator

2^x + 2^x

equals

2 \cdot 2^x

, which simplifies to

2^{x+1}

.

Adding two identical quantities is equivalent to multiplying that quantity by 2.

3

Convert the numerator to base 2 and simplify the fraction.

Since

4^{x+1} = (2^2)^{x+1} = 2^{2x+2}

, the fraction becomes

\frac{2^{2x+2}}{2^{x+1}} = 2^{(2x+2)-(x+1)} = 2^{x+1}

.

Converting all terms to a common base allows using the exponent quotient rule

a^m / a^n = a^{m-n}

.

4

Solve for

x

by equating the simplified power to 512.

Setting

2^{x+1} = 512 = 2^9

yields

x + 1 = 9

, so

x = 8

.

When exponential expressions with the same positive base (other than 1) are equal, their exponents must be equal.

Anahtar Kavram

Combining repeated addition of exponential terms and converting powers to a common base using exponent laws (

a^m \cdot a^n = a^{m+n}

and

\frac{a^m}{a^n} = a^{m-n}

).

Soru 22Soru →

If $x$ is a positive real number such that $x^{\frac{3}{4}} = 27$ , what is the value of $x^{-\frac{1}{2}}$ ?

\frac{1}{9}

9

-\frac{1}{9}

\frac{1}{3}

\frac{1}{27}

Cevabı ve açıklamayı göster

Cevap: $\frac{1}{9}$

Cevap

\frac{1}{9}

Raising both sides of

x^{\frac{3}{4}} = 27

to the power of

\frac{4}{3}

gives

x = (3^3)^{\frac{4}{3}} = 3^4 = 81

. Substituting

x = 81

into

x^{-\frac{1}{2}}

gives

81^{-\frac{1}{2}} = \frac{1}{\sqrt{81}} = \frac{1}{9}

.

Adım Adım Çözüm

1

Solve for

x

in the equation

x^{\frac{3}{4}} = 27

.

x = 27^{\frac{4}{3}} = (3^3)^{\frac{4}{3}} = 3^4 = 81

Raise both sides to the power of

\frac{4}{3}

to isolate

x

.

2

Evaluate

x^{-\frac{1}{2}}

for

x = 81

.

81^{-\frac{1}{2}} = \frac{1}{81^{\frac{1}{2}}} = \frac{1}{\sqrt{81}} = \frac{1}{9}

Apply the negative exponent rule

a^{-n} = \frac{1}{a^n}

and the fractional exponent rule

a^{\frac{1}{2}} = \sqrt{a}

.

Anahtar Kavram

Fractional and Negative Exponents

Tahmini Süre:1m 30s

Soru 23Soru →

If $x$ and $y$ are real numbers such that $x < 0 < y$ and $x^2 > y^2$ , which of the following statements must be true? Select all such statements.

Geçerli olan tümünü seçin

x^3 < y^3

\sqrt{x^2} + x = 0

x + y < 0

\sqrt{x^2 + y^2} = |x| + y

(-x)^2 = -x^2

Cevabı ve açıklamayı göster

Cevap: $x^3 < y^3$ ; $\sqrt{x^2} + x = 0$ ; $x + y < 0$

Cevap

The correct statements are

x^3 < y^3

,

\sqrt{x^2} + x = 0

, and

x + y < 0

.

The statement

x^3 < y^3

is correct because cubing a negative number keeps it negative while cubing a positive number keeps it positive. The statement

\sqrt{x^2} + x = 0

is correct because

\sqrt{x^2} = |x| = -x

for negative numbers. The statement

x + y < 0

is correct because

x^2 > y^2

implies

|x| > y

, meaning the negative component

x

has a larger absolute magnitude than the positive component

y

.

Adım Adım Çözüm

1

Analyze the signs of odd powers for

x < 0 < y

x^3 < 0

and

y^3 > 0

, which guarantees

x^3 < y^3

.

Odd powers preserve the original sign of the base.

2

Apply the definition of principal square roots to negative values

\sqrt{x^2} = |x| = -x

, so

\sqrt{x^2} + x = -x + x = 0

.

The square root symbol

\sqrt{}

denotes the principal (non-negative) root.

3

Compare absolute values using

x^2 > y^2

|x| > y \implies -x > y \implies x + y < 0

.

Since

x < 0

, its magnitude

|x|

is

-x

, which dominates the positive value

y

.

4

Evaluate the false options against exponent and radical rules

\sqrt{x^2 + y^2} \neq |x| + y

due to non-distributivity of roots, and

(-x)^2 = x^2 \neq -x^2

.

Radicals do not distribute over sums, and even powers eliminate negative signs.

Anahtar Kavram

Properties of real exponents, radical expressions, and absolute values for negative bases

Tahmini Süre:1m 30s

Soru 24Soru →

If $x = 4$ , what is the value of $\sqrt{9^x + 9^x + 9^x + 9^x}$ ?

81

162

324

648

1,296

Cevabı ve açıklamayı göster

Cevap: 162

Cevap

162

Combining the four identical terms under the square root gives

\sqrt{4 \cdot 9^4}

. Splitting the root using product rules yields

\sqrt{4} \cdot \sqrt{9^4} = 2 \cdot 9^2 = 2 \cdot 81 = 162

.

Adım Adım Çözüm

1

Substitute

x = 4

into the given radical expression.

The expression becomes

\sqrt{9^4 + 9^4 + 9^4 + 9^4}

.

Direct substitution of the given variable value.

2

Combine the four identical terms under the radical.

9^4 + 9^4 + 9^4 + 9^4 = 4 \cdot 9^4

.

Repeated addition of

4

identical terms is equivalent to multiplication by

4

.

3

Apply the product rule for square roots,

\sqrt{a \cdot b} = \sqrt{a} \cdot \sqrt{b}

.

\sqrt{4 \cdot 9^4} = \sqrt{4} \cdot \sqrt{9^4} = 2 \cdot 9^2.

Both

4

and

9^4

are perfect squares.

4

Evaluate the numerical value.

2 \cdot 81 = 162.

Simplifying the arithmetic expression.

Anahtar Kavram

Combining like terms under radical sign and applying product rules of square roots

Tahmini Süre:1m 15s

Soru 25Soru →

If $x$ and $y$ are real numbers such that $x < -1$ and $0 < y < 1$ , which of the following statements must be true? Select all such statements.

Geçerli olan tümünü seçin

x^2 > y^2

\sqrt{x^2 y^4} = -x y^2

x^3 > y^3

\sqrt{x^2} + y = x + y

(x + y)^2 = x^2 + y^2

Cevabı ve açıklamayı göster

Cevap: $x^2 > y^2$ ; $\sqrt{x^2 y^4} = -x y^2$

Cevap

The correct statements are the inequality asserting that the square of the first variable is greater than the square of the second, and the identity simplifying the square root of the product of the powers to negative the product of the first variable and the square of the second.

The statement comparing squared values is correct because any real number less than -1 has an absolute value greater than 1, so its square is strictly greater than 1, while any positive number less than 1 has a square strictly less than 1. The radical identity statement is correct because taking the square root of

x^2

yields

|x|

, which evaluates to

-x

when

x

is negative.

Adım Adım Çözüm

1

Analyze the given bounds for both variables.

For the first variable,

x < -1

, which implies

|x| > 1

,

x

is negative,

x^2 > 1

, and

x^3 < -1

. For the second variable,

0 < y < 1

, which implies

y > 0

,

y^2 < 1

, and

y^3 > 0

.

Establishing explicit bounds on magnitudes and signs is necessary to evaluate powers and absolute value roots.

2

Evaluate the inequality comparing the squared terms.

Since

x^2 > 1

and

y^2 < 1

, it follows directly that

x^2 > y^2

.

Transitive comparison across the threshold value of 1 proves the inequality holds.

3

Simplify the radical expression

\sqrt{x^2 y^4}

.

\sqrt{x^2 y^4} = \sqrt{x^2} \cdot \sqrt{y^4} = |x| \cdot y^2

. Since

x < 0

,

|x| = -x

, so the expression simplifies to

-x y^2

.

The principal square root of

x^2

must equal the absolute value

|x|

, which requires a sign flip when

x

is negative.

4

Verify remaining candidate expressions for potential fallacies.

Odd powers preserve negative signs so

x^3 < y^3

;

\sqrt{x^2}

equals

-x \neq x

; and expanding

(x+y)^2

produces a nonzero cross-term

2xy

.

Eliminating false options confirms that only two statements are universally true.

Anahtar Kavram

Principal square roots and even/odd power behaviors under negative variable bounds

Soru 26Soru →

If $m$ and $n$ are negative integers such that $m < n$ , which of the following statements must be true? Select all such statements.

Geçerli olan tümünü seçin

2^m < 2^n

\left(\frac{1}{2}\right)^m > \left(\frac{1}{2}\right)^n

m^2 < n^2

\sqrt{m^2} = m

2^{m+n} = 2^m + 2^n

Cevabı ve açıklamayı göster

Cevap: $2^m < 2^n$ ; \left(\frac{1}{2}\right)^m > \left(\frac{1}{2}\right)^n

Cevap

The statements

2^m < 2^n

and \left(\frac{1}{2}\right)^m > \left(\frac{1}{2}\right)^n must be true.

For the statement involving base

2

, since

2 > 1

, the exponential function is strictly increasing, so

m < n

guarantees

2^m < 2^n

. For the statement involving base

\frac{1}{2}

, since

0 < \frac{1}{2} < 1

, the function is strictly decreasing, meaning a smaller input

m

produces a larger output, so \left(\frac{1}{2}\right)^m > \left(\frac{1}{2}\right)^n.

Adım Adım Çözüm

1

Analyze the expression

2^m < 2^n

for base greater than 1

Since

b = 2 > 1

, raising

2

to a larger exponent yields a larger value. Because

m < n

,

2^m < 2^n

is always true.

Exponential functions with a base

b > 1

are strictly increasing.

2

Analyze the expression \left(\frac{1}{2}\right)^m > \left(\frac{1}{2}\right)^n for fractional base between 0 and 1

Since

b = \frac{1}{2}

is between

0

and

1

, raising

\frac{1}{2}

to a smaller exponent yields a larger value. Because

m < n

, \left(\frac{1}{2}\right)^m > \left(\frac{1}{2}\right)^n is always true.

Exponential functions with a base

0 < b < 1

are strictly decreasing.

3

Evaluate the remaining algebraic statements using counterexamples

For

m = -3

and

n = -2

:

m^2 = 9 > 4 = n^2

, so

m^2 < n^2

is false.

\sqrt{m^2} = \sqrt{9} = 3 \neq -3

, so

\sqrt{m^2} = m

is false.

2^{-5} = \frac{1}{32} \neq \frac{1}{8} + \frac{1}{4} = \frac{3}{8}

, so

2^{m+n} = 2^m + 2^n

is false.

A single counterexample disproves that a statement MUST be true.

Anahtar Kavram

Monotonicity of exponential functions and properties of square roots of negative bases

Soru 27Soru →

If $x$ is a real number such that $\frac{4^{x+2} - 4^x}{15} = 2^{x+3}$ , what is the value of $x$ ?

1

2

3

5

6

Cevabı ve açıklamayı göster

Cevap: 3

Cevap

3

Factoring out

4^x

from the numerator gives

4^x(4^2 - 1) = 15 \cdot 4^x

. Dividing by 15 simplifies the left-hand side to

4^x

. Expressing

4^x

as

2^{2x}

allows setting

2^{2x} = 2^{x+3}

. Equating the exponents

2x = x + 3

yields

x = 3

.

Adım Adım Çözüm

1

Factor the numerator of the left-hand side

4^{x+2} - 4^x = 4^x(4^2 - 1) = 4^x(16 - 1) = 15 \cdot 4^x

Factoring out the common exponential term

4^x

simplifies the subtraction.

2

Simplify the fraction on the left-hand side

\frac{15 \cdot 4^x}{15} = 4^x

The factor of 15 in the numerator cancels with 15 in the denominator.

3

Rewrite

4^x

in terms of base 2

4^x = (2^2)^x = 2^{2x}

Both sides must have a common base to equate their exponents.

4

Set the exponential expressions equal and solve for

x

2^{2x} = 2^{x+3} \implies 2x = x + 3 \implies x = 3

Since the bases are equal and positive (base 2), their exponents must be equal.

Anahtar Kavram

Factoring common exponential terms and equating exponents with identical bases

Tahmini Süre:1m 30s

Soru 28Soru →

If $n$ is a real number such that $\frac{27^n + 27^n + 27^n}{3^{n+2}} = 243$ , what is the value of $n$ ?

Cevabı ve açıklamayı göster

Cevap: 3

Cevap

3

Rewriting

27^n + 27^n + 27^n

as

3 \cdot (3^3)^n = 3^{3n+1}

allows the left-hand side to simplify to

\frac{3^{3n+1}}{3^{n+2}} = 3^{2n-1}

. Equating this to

243 = 3^5

gives

2n - 1 = 5

, which solves to

n = 3

.

Adım Adım Çözüm

1

Express repeated addition in the numerator as multiplication.

27^n + 27^n + 27^n = 3 \cdot 27^n

Adding three identical quantities is equivalent to multiplying one quantity by 3.

2

Convert base 27 to base 3 and apply exponent multiplication.

3 \cdot (3^3)^n = 3^1 \cdot 3^{3n} = 3^{3n+1}

Since

27 = 3^3

, using the power rule

(a^b)^c = a^{bc}

and product rule

a^b \cdot a^c = a^{b+c}

converts the numerator to a single power of 3.

3

Simplify the fraction using the quotient rule of exponents.

\frac{3^{3n+1}}{3^{n+2}} = 3^{(3n+1) - (n+2)} = 3^{2n-1}

Dividing exponential terms with the same base requires subtracting the exponent in the denominator from the exponent in the numerator.

4

Rewrite 243 with base 3 and equate exponents across the equal sign.

3^{2n-1} = 3^5 \implies 2n - 1 = 5

Since

243 = 3^5

, two exponential expressions with the same base are equal if and only if their exponents are equal.

5

Solve the linear equation for

n

.

2n = 6 \implies n = 3

Adding 1 to both sides yields

2n = 6

, and dividing by 2 yields

n = 3

.

Anahtar Kavram

Combining repeated addition of exponential terms and converting expressions to a common base using exponent rules.

Exponents, Powers, and Square Roots

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