# Logorithms on a first level learning Solve log_(5x−1)4 = 1/3

Logorithms on a first level learning
Solve log${}_{5x-1}$ $4$$=$$1/3$
$\left(5x-1{\right)}^{1/3}$=4
$\left(\left(5x-1{\right)}^{1/3}{\right)}^{3}$ = ${4}^{3}$
$5x-1=64$
$5x=65$
$13$
I am not sure where to go with this. I learned some things about logs before my class ended for the year. I just wanted to expand on my knowledge. This is more advance than what I am use to doing. Can someone please show me.
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GrEettarim3
If I didn't understand all of the steps, I would try to include all steps. One step that is missing is explicitly raising the base to the power of the LHS and RHS respectively.
${\mathrm{log}}_{5x-1}4=1/3$
$\left(5x-1{\right)}^{{\mathrm{log}}_{5x-1}4}=\left(5x-1{\right)}^{1/3}$
We know that ${a}^{{\mathrm{log}}_{a}Y}=Y$, similarly our equation becomes
$4=\left(5x-1{\right)}^{1/3}$
Now if $4=\left(5x-1{\right)}^{1/3}$ then
$4\cdot 4\cdot 4=\left(5x-1{\right)}^{1/3}\cdot \left(5x-1{\right)}^{1/3}\cdot \left(5x-1{\right)}^{1/3}$
${4}^{3}=\left(5x-1{\right)}^{1/3+1/3+1/3}$
$64=\left(5x-1{\right)}^{1}=5x-1$

deiluefniwf
It is easier if after the second line you cube both sides.