Question

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

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 (\(a^m = a^n\) implies \(m = n\) for \(a>0,a
eq1\)), we set \(8x = 3\).

Step4: Solve for x

Divide both sides by \(8\): \(x=\frac{3}{8}\)? Wait, no, wait. Wait, I made a mistake in step 1. Wait, the original equation is \(4^{4x}=8\)? Wait, no, looking back, the original equation is \(4^{4x}=8\)? Wait, no, the user wrote \(4^{4x}=8\)? Wait, no, the image shows \(4^{4x}=8\)? Wait, no, maybe it's \(4^{x}=8\)? Wait, no, the user's image: let me check again. The equation is \(4^{4x}=8\)? Wait, no, the options: A is 2, B is 3/8, C is 8/3, D is 3/4. Wait, maybe I misread the exponent. Wait, maybe it's \(4^{x}=8\)? No, the exponent is \(4x\)? Wait, no, let's re - do.

Wait, correct approach:

Given \(4^{4x}=8\)

Rewrite \(4 = 2^2\) and \(8=2^3\)

So \((2^2)^{4x}=2^3\)

\(2^{8x}=2^3\)

Since the bases are equal, exponents are equal: \(8x = 3\)

\(x=\frac{3}{8}\). Wait, but option B is \(\frac{3}{8}\). Wait, maybe the original equation was \(4^{x}=8\)? No, the exponent is \(4x\) in the image. Wait, let's check the options. Option B is \(\frac{3}{8}\), option A is 2, option C is \(\frac{8}{3}\), option D is \(\frac{3}{4}\).

Wait, maybe I misread the exponent. Let's assume the equation is \(4^{x}=8\) (maybe a typo in the exponent). Let's try that.

If \(4^{x}=8\), rewrite \(4 = 2^2\), \(8 = 2^3\), so \((2^2)^{x}=2^3\), \(2^{2x}=2^3\), then \(2x = 3\), \(x=\frac{3}{2}\), which is not an option.

Wait, going back to the original problem: the equation is \(4^{4x}=8\). Let's solve it correctly.

\(4^{4x}=8\)

\((2^2)^{4x}=2^3\)

\(2^{8x}=2^3\)

So \(8x = 3\)

\(x=\frac{3}{8}\), which is option B.

Answer:

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