Solve log$_{5x-1}$ $4$ $=$ $1/3$
$(5x-1)^{1/3}$=4
$((5x-1)^{1/3})^3$ = $4^3$
$5x-1=64$
$5x=65$
$13$
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.
On
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. $$\log_{5x-1}4 = 1/3$$ $$(5x-1)^{\log_{5x-1}4} = (5x-1)^{1/3}$$
We know that $a^{\log_aY}=Y$, similarly our equation becomes
$$4 = (5x-1)^{1/3}$$ Now if $4 = (5x-1)^{1/3}$ then $$4\cdot4\cdot4 = (5x-1)^{1/3}\cdot (5x-1)^{1/3}\cdot (5x-1)^{1/3}$$ $$4^3 = (5x-1)^{1/3+1/3+1/3}$$ $$64 = (5x-1)^{1} = 5x-1$$
If you need a little refresher on the rules of exponents, try looking at http://mathontrack.comze.com/exponentials2.html
It is easier if after the second line you cube both sides.