If \left(\frac{1}{8}\right)^{2x - 3} = 16^{x - 2}, what is the value of x? — SAT

If $\left(\frac{1}{8}\right)^{2x - 3} = 16^{x - 2}$ , what is the value of $x$ ?

Cevap: 1.7

1.7 (or 17/10)

The correct answer is 1.7 (or 17/10). By writing

\frac{1}{8}

2^{-3}

and

16

2^4

, the equation becomes

(2^{-3})^{2x - 3} = (2^4)^{x - 2}

. Applying exponent rules yields

2^{-6x + 9} = 2^{4x - 8}

. Equating the exponents gives

-6x + 9 = 4x - 8

, which simplifies to

10x = 17

, resulting in

x = 1.7

Express both bases as powers of

2

(2^{-3})^{2x - 3} = (2^4)^{x - 2}

To solve the exponential equation, both sides must have a common base. Both

\frac{1}{8}

and

16

can be rewritten as powers of

2

Apply the power-of-a-power exponent rule

(a^m)^n = a^{mn}

to simplify the exponents.

2^{-3(2x - 3)} = 2^{4(x - 2)}

, which simplifies to

2^{-6x + 9} = 2^{4x - 8}

This allows us to write each side as a single exponential term with a base of

2

Set the exponents equal to each other.

-6x + 9 = 4x - 8

Since the bases are equal, their exponents must be equal for the equation to hold.

Solve the linear equation for

x

10x = 17

, which gives

x = 1.7

Isolate the variable

x

by algebraic manipulation.

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

Alternatively, you can apply the definition of logarithms to both sides of the equation. Taking the logarithm base 2 of both sides gives

\log_2\left(\left(\frac{1}{8}\right)^{2x-3}\right) = \log_2\left(16^{x-2}\right)

. Using the power property of logarithms, this simplifies to

(2x - 3)\log_2\left(2^{-3}\right) = (x - 2)\log_2\left(2^4\right)

, which simplifies directly to

-3(2x - 3) = 4(x - 2)

. Solving this linear equation yields the same result,

x = 1.7

Tahmini Süre:1m 30s

Soru