Question

solve for x:
$32^{x + 4} = \left(\frac{1}{128}\
ight)^{-4x - 15}$

Explanation:

Step1: Express bases as powers of 2

We know that \(32 = 2^5\) and \(128 = 2^7\), so \(\frac{1}{128}=2^{-7}\). Substitute these into the equation:
\((2^5)^{x + 4}=(2^{-7})^{-4x - 15}\)

Step2: Apply exponent rules (\((a^m)^n=a^{mn}\))

Simplify both sides using the power - of - a - power rule:
For the left - hand side: \((2^5)^{x + 4}=2^{5(x + 4)}=2^{5x+20}\)
For the right - hand side: \((2^{-7})^{-4x - 15}=2^{(-7)\times(-4x - 15)}=2^{28x + 105}\)

Step3: Set exponents equal (since \(a^m=a^n\Rightarrow m = n\) for \(a>0,a

eq1\))
Since the bases are the same (\(a = 2>0,a
eq1\)), we can set the exponents equal to each other:
\(5x+20 = 28x+105\)

Step4: Solve for x

Subtract \(5x\) from both sides:
\(20=28x - 5x+105\)
\(20 = 23x+105\)
Subtract 105 from both sides:
\(20-105 = 23x\)
\(- 85=23x\)
Divide both sides by 23:
\(x=-\frac{85}{23}\)

Answer:

\(x =-\frac{85}{23}\)