Solving $4^{32^x} = 16^{8^x}$

Question

Solve for $x$: $4^{32^x} = 16^{8^x}$.

So I have tried using log (so $x\log{4^{32}} = x\log{16^8}$), but that wasn't very helpful to me, and some random guessing gave me an answer of $1/2$, but I was wondering how I could be more mathematical…

Answer 1

$$4^{32^x}=(4^2)^{8^x}$$

$$32^x=2 \cdot 8^x$$

$$2^{5x}=2^{1+3x}$$

$$5x=1+3x$$

$$x=\frac12$$

Answer 2

Since$$ 4^{32^x} = 2^{2 \cdot 32^x} = 2^{2 \cdot 2^{5x}} = 2^{2^{5x + 1}} $$ and$$ 16^{8^x} = 2^{4 \cdot 8^x} = 2^{4 \cdot 2^{3x}} = 2^{2^{3x + 2}}, $$ then$$ 4^{32^x} = 16^{8^x} \Longleftrightarrow 2^{2^{5x + 1}} = 2^{2^{3x + 2}} \Longleftrightarrow 2^{5x + 1} = 2^{3x + 2}\\\Longleftrightarrow 5x + 1 = 3x + 2 \Longleftrightarrow x = \frac{1}{2}. $$

Answer 3

If you take logarithm, you get $$ 32^x\,\log 4=8^x\,\log 16=2\times 8^x\,\log 4. $$ So $$ 2=\frac{32^x}{8^x}=\left(\frac{32}8\right)^x=4^x. $$ Now you either note directly that $x=1/2$, or take log again to get $$ \log 2 =x\,\log4=2x\log 2, $$ or $$ 2x=1.$$