Solving an equation involving $\log_{10}$

Question

If $$\log_{10}(x)\log_{10}(2) = 2$$ What is $x$ ?

WolframAlpha says $x = e^{\frac2{\log_{10}(2)}}$

But i don't understand why it is.. Please explain it. Thanks

Answer 1

There's really not much to it, but you probably used $\ln$ when asking W|A (log is interpreted as the natural logaritm by default). $$\log_{10} x \log_{10} 2 = 2 \\ \Rightarrow \log_{10} x = \frac2{\log_{10}2} \\ \Rightarrow x = 10^{\log_{10} x} = 10^{\frac2{\log_{10}2}}$$

Answer 2

That's because, when you write down $\log$ in WolframAlpha, it is interpreted as $\log_e$ or $\ln$. So WolframAlpha solved the equation

$$\ln(x)\ln(2)=2$$

Which is solved by first dividing the equation by $\ln 2$, obtaining $$ln(x) = \frac{2}{\ln 2}$$

Then using the fact that $a=b\iff e^a=e^b$ to get

$$(x=)e^{\ln x} = e^{\frac{2}{\ln2}}$$

Answer 3

It's just a rearrangement of variables.

So, say:

• a = log(x)
• b = log(2)
• c = 2

Then we have a*b = c, which means a=c/b

To get x by itself simply take the exponential of both sides (because the exponential of the log of x is just x).