$\log x$ and $\ln x$ comparison

Question

If $0 < x < 1$, then compare the values $\log x$ and $\ln x$.

I tried to take advantage of the monotonicity of the functions $f(x)=\log x$ and $g(x)=\ln x$, but nothing. Any thoughts?

Answer 1

Recall property of logarithms : $$\log_{n}m =\frac{\log_{a}m}{\log_{a}n}$$

Therefore; $\ln n = \log_e n = \dfrac{\log_{10} n}{\log_{10} e} \implies \dfrac{\ln n}{\log_{10} n}= \dfrac{1}{\log_{10}e} = \ln 10$

Hence : $$\color{blue}{\ln n = \log_{10}n \cdot \ln 10}$$

Answer 2

Hint: $\log_b x = \frac{\log_c x}{\log_c b}$ for any $b, c, x > 0$.