How to solve $\log(1+ax)=b\log(1+x)?$

Question

I want to solve this equation for $x$:

$\log(1+ax)=b\log(1+x)$

In this equation $a,b>1$. Would you please help how should I do that or what should I study?

Thanks in advance.

Answer 1

Well, we know that:

$$\ln\left(x^y\right)=y\ln\left(x\right)\tag1$$

So:

$$\ln\left(1+ax\right)=b\ln\left(1+x\right)=\ln\left(\left(1+x\right)^b\right)\tag2$$

Now, taking $\exp$ on both sides leads to:

$$1+ax=\left(1+x\right)^b\tag3$$

Now, you can see the problem why it is not easy to solve for a general $x$.

Answer 2

Hint:

$n\cdot \log x = \log x^n$