Solve equation with logarithm base 10

I am going back to study log and unfortunately I don't know a lot. I need to solve this: $$100= 10\log_{10} \left(50/x\right)$$ I did the wrong calculation just moving stuff to the left, but I've been told is not right: \begin{align} 100 \cdot x &= 10\log_{10} \cdot 50\\ x &= \left(10\log_{10} \cdot 50\right) / 100\\ x &= 0.016 \end{align} I know that it is wrong, can someone explain me how to solve this?

• Could you use brackets in your formulas? It's not clear if you mean $^{10}\log(10)\cdot\frac{50}x$, $^{10}\log\left(10\cdot\frac{50}x\right)$ or $10\log_{10}\left(\frac{50}x\right)$. I suppose it's the third one. – punctured dusk Mar 4 '14 at 9:57
• yes it's the third one barto, thanks a lot, sorry I am learning to use the site – jsab Mar 4 '14 at 9:59

$x\ne 0$ is part of the argument of $\log_{10},\,$ you cannot move it out the way you did. You can use $\log_{10}(50/x)= \log_{10}50 -\log_{10}x,\,$ or something like this: $$100 = 10\log_{10}\left(\frac{50}{x}\right)\quad (x\ne0)$$ $$\iff 10 = \log_{10}\left(\frac{50}{x}\right)$$ $$\iff 10^{10}=\frac{50}{x}$$ $$\iff x=\frac{50}{10^{10}}=\frac{1}{200000000}=5\times10^{-9}$$
The mistake is that we can't just pull the factor $\frac1x$ out of the logarithm, i.e, $$\log_{10}\left(\frac{50}x\right)\neq\log_{10}(50)\cdot\frac1x$$ in general. The logarithm (in particular the base-$10$ logarithm) has no rule like $$\log(a\cdot b)=\log(a)\cdot\log(b).$$ (In fact we have $\log ab=\log a+\log b$.)
What we can use here is: $$\log_{10}(a)=b\quad\Longleftrightarrow\quad a=10^b.$$
Before making this possible, we need a small manipulation to the given equation: $$100=10\cdot\log_{10}\left(\frac{50}x\right)\quad\Longleftrightarrow\quad10=\log_{10}\left(\frac{50}x\right).$$ Now we can apply the above rule, with $a=\frac{50}x$ and $b=10$: $$\frac{50}x=a=10^b=10^{10}.$$ Look, the logarithm has disappeared! From here we can conclude $$x=\frac{50}{10^{10}}=\frac1{200000000}.$$