Question

5 solve the exponential equation.
$4^{4x} = 8$
a $\frac{8}{3}$
b $\frac{3}{4}$
c $2$
d $\frac{3}{8}$

Explanation:

Step1: Express bases as powers of 2

Rewrite \(4\) as \(2^2\) and \(8\) as \(2^3\). So the equation \(4^{4x}=8\) becomes \((2^2)^{4x}=2^3\).

Step2: Simplify exponents

Using the power - of - a - power rule \((a^m)^n=a^{mn}\), we have \(2^{8x}=2^3\).

Step3: Set exponents equal

Since the bases are the same and the exponential function \(y = 2^x\) is one - to - one, we can set the exponents equal to each other: \(8x = 3\).

Step4: Solve for x

Divide both sides of the equation \(8x = 3\) by \(8\): \(x=\frac{3}{8}\). Wait, no, wait. Wait, \((2^2)^{4x}=2^{8x}\), and \(8 = 2^3\), so \(2^{8x}=2^3\) implies \(8x = 3\), so \(x=\frac{3}{8}\)? Wait, no, I made a mistake. Wait, \(4 = 2^2\), so \(4^{4x}=(2^2)^{4x}=2^{8x}\), and \(8 = 2^3\). So \(2^{8x}=2^3\), so \(8x=3\), so \(x = \frac{3}{8}\)? But let's check again. Wait, maybe I messed up the exponent. Wait, \(4^{4x}\): \(4\) is \(2^2\), so \((2^2)^{4x}=2^{2\times4x}=2^{8x}\), and \(8 = 2^3\). So if \(2^{8x}=2^3\), then \(8x = 3\), so \(x=\frac{3}{8}\). But let's check the options. Option D is \(\frac{3}{8}\). Wait, but let's verify by plugging back in. If \(x = \frac{3}{8}\), then \(4^{4\times\frac{3}{8}}=4^{\frac{3}{2}}\). \(4^{\frac{3}{2}}=(2^2)^{\frac{3}{2}}=2^{3}=8\), which matches the right - hand side. So the correct value of \(x\) is \(\frac{3}{8}\).

Answer:

D. \(\frac{3}{8}\)