Question

1. solve the equation.

$64^{x} = \left(\frac{1}{4}\
ight)^{2x + 15}$
$x = \square$

Explanation:

Step1: Express bases as powers of 2

We know that \(64 = 2^6\) and \(\frac{1}{4}=2^{-2}\). So we rewrite the equation:
\((2^6)^x=(2^{-2})^{2x + 15}\)

Step2: Apply exponent rules

Using the power - of - a - power rule \((a^m)^n=a^{mn}\), we get:
\(2^{6x}=2^{-2(2x + 15)}\)

Step3: Simplify the exponents

Simplify the right - hand side exponent: \(-2(2x + 15)=-4x-30\). So the equation becomes \(2^{6x}=2^{-4x - 30}\)

Step4: Set exponents equal

Since the bases are the same (\(a^m=a^n\) implies \(m = n\) for \(a>0,a
eq1\)), we set the exponents equal to each other:
\(6x=-4x - 30\)

Step5: Solve for x

Add \(4x\) to both sides of the equation:
\(6x + 4x=-4x-30 + 4x\)
\(10x=-30\)
Divide both sides by 10:
\(x=\frac{-30}{10}=-3\)

Answer:

\(x = - 3\)