A question about $\log_{10} ⁡(x) = k\ln⁡x $

Question

I need help on this question:

$\log_{10} ⁡(x) = k\ln⁡x $

By raising $10$ to the power of both sides, show that $k= \frac1{\ln⁡10}$ .

I have absolutely no clue on how to start.

Answer 1

By definition $10^{\log_{10}(x)}= x$

What is $10^{k\ln(x)}$? Write $10=e^{\ln(10)}$, then we get $(e^{\ln(10)})^{k \ln(x)} = e^{\ln(x) k \ln(10)}= (e^{\ln(x)})^{k \ln(10)}= x^{k \ln(10)}$, which thus equals $x=x^1$, hence $k\ln(10)=1$ etc..

Answer 2

I sometimes get confused about these logarithm questions - which way round the logs are and whether to divide or multiply.

If in doubt, I generally go back to powers.

So suppose $x=10^a$ and $10=e^b$ then $x=(e^b)^a=e^{ab}$

Then I can take logarithms and see what is going on in that language.