Rapsinincke

2022-07-08

logarithms equations, different bases
solve equations: ${\mathrm{log}}_{x}10+2{\mathrm{log}}_{10x}10-3{\mathrm{log}}_{100x}10=0$ so I tried to use ${\mathrm{log}}_{a}b=\frac{1}{{\mathrm{log}}_{b}a}$ but it didn't work for me.

Oliver Shepherd

Expert

${\mathrm{log}}_{10x}10=\frac{1}{{\mathrm{log}}_{10}\left(10x\right)}=\frac{1}{1+{\mathrm{log}}_{10}x}=\frac{1}{1+u}$
and similarly
${\mathrm{log}}_{100x}10=\frac{1}{{\mathrm{log}}_{10}\left(100x\right)}=\frac{1}{2+{\mathrm{log}}_{10}x}=\frac{1}{2+u}.$
So you have
$\frac{1}{u}+\frac{2}{1+u}-\frac{3}{2+u}=0.$
If you multiply both sides by $u\left(1+u\right)\left(2+u\right)$, you get
$\left(1+u\right)\left(2+u\right)+2u\left(2+u\right)-3u\left(1+u\right)=0.$
Multiply that out, then collect like terms, then you have a quadratic equation.

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