If 4^{2n-3} = \left(\frac{1}{8}\right)^{n-5}, what is the value of n? — SAT

If $4^{2n-3} = \left(\frac{1}{8}\right)^{n-5}$ , what is the value of $n$ ?

Cevap: 3

Rewriting the bases 4 and

\frac{1}{8}

as powers of 2 gives

(2^2)^{2n-3} = (2^{-3})^{n-5}

. Applying the exponent power rule simplifies this to

2^{4n-6} = 2^{-3n+15}

. Since the bases are equal, the exponents must be equal:

4n-6 = -3n+15

. Solving this linear equation by adding

3n

and 6 to both sides gives

7n = 21

, which yields

n = 3

Express the bases 4 and

\frac{1}{8}

as powers of the prime base 2.

4 = 2^2

and

\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.

Substitute these bases back into the original equation.

(2^2)^{2n-3} = (2^{-3})^{n-5}

This sets up the equation to simplify the exponent terms.

Apply the power of a power rule,

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

, by multiplying the exponents on both sides.

2^{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.

Equate the exponents since the bases are now equal, and solve for

n

4n-6 = -3n+15 \implies 7n = 21 \implies n = 3

b^x = b^y

where

b > 0

and

b \neq 1

, then

x = y

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

Soru