2022-06-21

Logarithmic functions?
I am stuck on this question which as follows:
$\mathrm{log}\left(x\right)+\mathrm{log}\left(x-3\right)=\mathrm{log}\left(10x\right)$
I have tried the following and not sure if I am doing it correctly...
1)
$\mathrm{log}\left(x\right)+\mathrm{log}\left(x-3\right)=\mathrm{log}\left(10x\right)\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\mathrm{log}\left(\left(x\right)\left(x-3\right)\right)=\mathrm{log}\left(10x\right)$
$\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\mathrm{log}\left({x}^{2}-3x\right)=\mathrm{log}\left(10x\right)\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{x}^{2}-3x=10x$
${x}^{2}-13x=0\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}x=0,x=13$
2)
$\mathrm{log}\left(x\right)+\mathrm{log}\left(x-3\right)=\mathrm{log}\left(10x\right)$
$\mathrm{log}\left(x\right)+\mathrm{log}\left(x\right)-\mathrm{log}\left(3\right)=\mathrm{log}\left(10x\right)$
$x+x-3=10x\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}x=-3/8$
Are these approaches on the right path?

Jake Mcpherson

The first approach is correct. The second is not: $\mathrm{log}\left(x-3\right)\ne \mathrm{log}x-\mathrm{log}3$
The next step is to check for extraneous solutions, by verifying which solutions ( $x=0$ and $x=13$) satisfy the original equation.

fabios3

I am not sure if $log\left(z\right)$ means $lo{g}_{10}\left(z\right)$ or $lo{g}_{e}\left(z\right)$ i.e $ln\left(z\right)$ so I will go with convention and assume $log\left(z\right)\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}lo{g}_{10}\left(z\right)$
Here goes:
so $x\in \left\{0,13\right\}$
But for $x=0$ the original equation becomes: $log\left(0\right)+log\left(-3\right)=log\left(0\right)$ his is a problem because $log\left(z\right)$ is defined only for $0 if z is real.
Logarithms of complex numbers quickly get very messy. So confining the arguments of the log functions to positive reals leaves a single answer: $x=13$
I hope this was not too much detail.

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