$(5^{2x}-1)(5^x)=1/5^x$ solve

Question

I have the problem $(5^{2x}-1)(5^x) = 1/5^x$.

I have already simplified it to $5^{3x}-1=1/5^x$ My question is when I do $\log$ base $5$ to the left side of the equation to get $3x-1$ by itself so that I can figure out $x$, what do I get when I do $\log$ base $5$ to the other side of the equation? Does the $5$ cancel out? What happened to the exponent etc.

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Number 2 is the problem i am talking about in the image above

all these are confusing me, all I want to know is when I do log base 5 to one over 5^x (1/5^x), what happens :(

Answer 1

Hint: let $y = 5^x$. then you have: $y^3 - y = 1/y$. You can take it from here.

Answer 2

Let $a=5^x$. Then, $a^3-a=\dfrac{1}{a}$, so $$a^4-a^2-1=0$$ The solutions are $$a=\pm 1.2720196495139988,\pm0.7861513777573095i$$ Take the $\log$ base $5$ to get (note - only the $\log$s of the positive solutions are taken) $$x= 0.14949685892,(-1.2030295627\times10^{-1})+\dfrac{i\pi}{2}$$ As @RecklessReckoner has said, we will consider only the real values to get $$\boxed{x= 0.14949685892\mbox{, and as said in the comments, }\log_5\left(\sqrt{\dfrac{1+\sqrt{5}}{2}}\right)}$$

Answer 3

Firstly, your simplification of the lefthand side is incorrect. Note that: $$(5^{2x}-1)(5^x)\ne 5^{3x-1}$$ And instead: $$(5^{2x}-1)(5^x)=(5^{2x}\cdot 5^x)-(1\cdot5^x)=5^{2x+x}-5^x=5^{3x}-5^x$$ This is by: $$(a-b)c=ac-bc$$

Now you'll have $$5^{3x}-5^x=\frac{1}{5^x}$$ To solve, let $y=5^x$ as was suggested in E-theory's answer, and you'll get: $$y^3-y=\frac{1}{y}$$ Also note that:$$5^{3x}=\left(5^x\right)^3$$