Summer Bradford

2022-06-21

Logarithmic functions?

$\mathrm{log}(x)+\mathrm{log}(x-3)=\mathrm{log}(10x)$

I have tried the following and not sure if I am doing it correctly...

1)

$\mathrm{log}(x)+\mathrm{log}(x-3)=\mathrm{log}(10x)\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\mathrm{log}((x)(x-3))=\mathrm{log}(10x)$

$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\mathrm{log}({x}^{2}-3x)=\mathrm{log}(10x)\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{x}^{2}-3x=10x$

${x}^{2}-13x=0\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}x=0,x=13$

2)

$\mathrm{log}(x)+\mathrm{log}(x-3)=\mathrm{log}(10x)$

$\mathrm{log}(x)+\mathrm{log}(x)-\mathrm{log}(3)=\mathrm{log}(10x)$

$x+x-3=10x\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}x=-3/8$

Are these approaches on the right path?

Jake Mcpherson

Beginner2022-06-22Added 23 answers

The initial strategy is the best one. Not the second is: $\mathrm{log}(x-3)\ne \mathrm{log}x-\mathrm{log}3$

The following stage is to confirm which solutions are necessary to eliminate any unnecessary ones ( $x=0$ and $x=13$) satisfy the original equation.

fabios3

Beginner2022-06-23Added 10 answers

I am not sure if $log(z)$ means $lo{g}_{10}(z)$ or $lo{g}_{e}(z)$ i.e $ln(z)$ so I will go with convention and assume $log(z)\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}lo{g}_{10}(z)$

Here goes:

so $x\in \{0,13\}$

But for $x=0$ the original equation becomes: $log(0)+log(-3)=log(0)$ his is a problem because $log(z)$ is defined only for $0z$ if z is real.

Complex number logarithms quickly get disorganized. The only response is left when the log function arguments are limited to positive reals: $x=13$

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