Question

what is the true solution to the logarithmic equation below?
$log_{4}log_{4}(2x) = 1$
$\bigcirc$ $x = 2$
$\bigcirc$ $x = 8$
$\bigcirc$ $x = 64$
$\bigcirc$ $x = 128$

Explanation:

Step1: Solve the outer logarithm

Using the definition of logarithms, if $\log_{a}b = c$, then $b = a^{c}$. For $\log_{4}[\log_{4}(2x)] = 1$, we have $\log_{4}(2x)=4^{1}=4$.

Step2: Solve the inner logarithm

Now, for $\log_{4}(2x) = 4$, using the logarithm definition again, we get $2x = 4^{4}$.

Step3: Calculate the value of \(x\)

First, calculate $4^{4}=256$. Then, from $2x = 256$, divide both sides by 2: $x=\frac{256}{2}=128$? Wait, no, wait, 4^4 is 256? Wait, no, wait, in step 1, when we have $\log_{4}[\log_{4}(2x)] = 1$, so the outer log: let $y = \log_{4}(2x)$, then $\log_{4}y = 1$, so $y = 4^1 = 4$. Then, $\log_{4}(2x)=4$, so $2x = 4^4$? Wait, no, 4^4 is 256? Wait, no, 4^2 is 16, 4^3 is 64, 4^4 is 256? Wait, but let's check again. Wait, no, if $\log_{4}(2x)=4$, then $2x = 4^4$? Wait, no, the base is 4, so $\log_{b}a = c$ implies $a = b^c$. So $\log_{4}(2x)=4$ implies $2x = 4^{4}$? Wait, 4^4 is 256, then 2x=256, x=128? But wait, let's check the options. Wait, the options are x=2,8,64,128. Wait, maybe I made a mistake. Wait, no, wait, step 1: $\log_{4}[\log_{4}(2x)] = 1$ => $\log_{4}(2x)=4^1=4$. Then step 2: $\log_{4}(2x)=4$ => $2x=4^4$? Wait, no, 4^4 is 256? Wait, no, 4^4 is 256? Wait, 4^1=4, 4^2=16, 4^3=64, 4^4=256. Then 2x=256, x=128. But let's check with x=128: $\log_{4}[\log_{4}(2*128)]=\log_{4}[\log_{4}(256)]$. $\log_{4}(256)$: since 4^4=256, so $\log_{4}(256)=4$. Then $\log_{4}(4)=1$, which matches the equation. Wait, but let's check x=64: 2x=128, $\log_{4}(128)$. What's $\log_{4}(128)$? 4^3=64, 4^4=256, so 128 is between 4^3 and 4^4. So $\log_{4}(128)=\log_{4}(4^{3.5})=3.5$, then $\log_{4}(3.5)$ is not 1. So x=128: 2x=256, $\log_{4}(256)=4$, then $\log_{4}(4)=1$. Yes, that works. Wait, but let's check the options. Wait, the options include x=128. But wait, maybe I made a mistake in the exponent. Wait, no, let's re-express. Wait, maybe I messed up the exponent. Wait, step 1: $\log_{4}[\log_{4}(2x)] = 1$ => $\log_{4}(2x)=4^1=4$. Then step 2: $\log_{4}(2x)=4$ => $2x=4^4$? Wait, no, 4^4 is 256, but 4^2 is 16, 4^3 is 64, 4^4 is 256. Wait, but let's check x=64: 2x=128, $\log_{4}(128)$. Let's compute $\log_{4}(128)$. Since 4=2^2, 128=2^7. So $\log_{4}(128)=\log_{2^2}(2^7)=\frac{7}{2}=3.5$. Then $\log_{4}(3.5)$ is not 1. x=8: 2x=16, $\log_{4}(16)=2$, then $\log_{4}(2)\approx0.5$, not 1. x=2: 2x=4, $\log_{4}(4)=1$, then $\log_{4}(1)=0$, not 1. x=128: 2x=256, $\log_{4}(256)=4$ (since 4^4=256), then $\log_{4}(4)=1$, which matches. Wait, but earlier I thought 4^4 is 256, which is correct. So x=128. Wait, but let's check the options again. The options are x=2,8,64,128. So the correct answer is x=128? Wait, but let's re-examine the steps. Wait, maybe I made a mistake in step 2. Wait, $\log_{4}(2x)=4$: so 2x=4^4? Wait, no, 4^4 is 256, but 4^2 is 16, 4^3 is 64, 4^4 is 256. Wait, but 4^4 is 256, so 2x=256, x=128. Yes, that's correct.

Wait, but wait, let's check the calculation again. Step 1: $\log_{4}[\log_{4}(2x)] = 1$. By the definition of logarithm, if $\log_{a}b = c$, then $b = a^c$. So here, base a=4, b=log₄(2x), c=1. So log₄(2x) = 4¹ = 4. Then, for log₄(2x) = 4, again using the logarithm definition, 2x = 4⁴. 4⁴ is 256. Then 2x = 256 => x = 256 / 2 = 128. So x=128 is the solution.

Answer:

x = 128 (the option with x=128)