for what value of x does $4^{x}=\\left(\\frac{1}{8}\\right)^{x + 5}$?\n-15\n-3\n3\n15
for what value of x does $4^{x}=\\left(\\frac{1}{8}\\right)^{x + 5}$?\n-15\n-3\n3\n15
Answer
Explanation:
Step1: Rewrite bases as powers of 2
$4^x=(2^2)^x = 2^{2x}$ and $\left(\frac{1}{8}\right)^{x + 5}=(2^{-3})^{x+5}=2^{-3(x + 5)}=2^{-3x-15}$.
Step2: Set exponents equal
Since $4^x=\left(\frac{1}{8}\right)^{x + 5}$, then $2^{2x}=2^{-3x - 15}$. So, $2x=-3x-15$.
Step3: Solve for x
Add $3x$ to both sides: $2x+3x=-3x - 15+3x$, which gives $5x=-15$. Divide both sides by 5: $x=\frac{-15}{5}=-3$.
Answer:
-3