Multiplying two logarithms

Question

I've searched for some answer already, but couldn't find any solution to this problem. Apparently, there's no rule for the product of two logarithms. How would I then find the exact solution of this problem? $$ \log(x) = \log(100x) \, \log(2) $$

Answer 1

How would I then find the exact solution of this problem?

Manipulate the equation to isolate $x$. \begin{align*} \log(x) &= (\log(100)+\log(x))\log(2) \\ \log(x) &=\log(100)\log(2)+\log(x)\log(2)\\ \log(x)-\log(x)\log(2)&=\log(100)\log(2)\\ \log(x)(1-\log(2))&=\log(100)\log(2) \\ \log(x)&=\log(100)\log(2)/(1-\log(2))\\ \end{align*} Then resolve $x$ with whatever base your logarithm is using. E.g. with base 10, $$x\approx7.267$$

Answer 2

Fill in details after you check the basic properties of logarithms, and assuming $\;\log=\log_{10}\;$:

$$\log x=\log 100x\cdot\log2=\left(\log100+\log x\right)\log2\implies$$

$$(1-\log2)\log x=2\log2\implies\ldots$$

Answer 3

$\log(x) = (\log(100) + \log(x))\cdot\log2$

$\log(x) = \log(100)\cdot\log(2) + \log(x)\cdot\log(2)$

This is a linear equation in $\log(x)$!

Answer 4

The question does not specify the base $B$ of the logarithm, but it will affect the solution, so we make it explicit: \begin{align} \log_B(x) &= \log_B(100\, x) \, \log_B(2) \\ &= (\log_B(100) + \log_B(x)) \, \log_B(2) \iff \\ (1 - \log_B(2)) \log_B(x) &= \log_B(2) \log_B(100) \\ \end{align} For $B = 2$ the LHS vanishes and we have no solution, as the logarithms on the RHS do not vanish.

For $B \ne 2$ we can continue: \begin{align} \log_B(x) = \frac{\log_B(2) \, \log_B(100)}{1 - \log_B(2)} = f(B) \iff \\ x = B^{f(B)} = B^{(\log_B(2) \, \log_B(100))/(1 - \log_B(2))} \end{align}

For $B=e$ one gets $$ x = e^{f(e)} = e^{10.4025\dotsb} = 32944.48\dotsb $$

Here are graphs of $f(B)$:

(Links to larger versions: left, right)