Quick Log problem?

Question

$2^{(x^3)} = 3^{(x^2)}$

Solve for x

I'm pretty sure I use logs to solve this, but how? to what base? I'm kinda lost.. Thanks

Answer 1

Taking logs to any base, $$x^3\log2=x^2\log3\ ,$$ and therefore either $x=0$ or $x=(\log3)/(\log2)$. Note that by the "change of base" formula, the last expression is the same no matter what base you are using.

Answer 2

Take natural log $$ 2^{x^3} = 3^{x^2} \implies x^3 \ln 2 = x^2 \ln 3 \implies x^2 \left( x\ln2 - \ln 3\right) = 0 \implies x_{1,2} = 0,\ x_3 = \frac {\ln 3}{\ln 2} = \log_2 3 $$

Answer 3

Note the following facts:

1. $\log(a^{(b)}) = b\log(a)$. This is the key here (assuming $\log(a)$ exists).

2. When you raise a value to 0 power, you get a value of 1, so $x^0=5^0=1$

It is obvious that when $x=0$, the 2 sides would be equal.

Now the other possible solution, starting from:

$2^{(x^3)} = 3^{(x^2)}$

Apply the log function to both sides to get:

$\log(2^{(x^3)}) = \log(3^{(x^2)})$

$x^3\log(2) = x^2\log(3)$

assuming $x$ is not zero,

$x=\log(3)/\log(2)=1.58496250072$

Now an interesting point here would be why would get the same answer if you used another base for the log?