1. I have to solve:

(log(x^2) + log(x^4))/log(90x) = 3

I've been on it for 25 minutes and have gotten as far as:

log(x)/log(90x) = 1/2

And I'm not sure if I'm even going the right path lol, just trying to play around with properties of logs and not sure how to maneuver this. Thanks for any help

2. Originally Posted by DannyMath
I have to solve:

(log(x^2) + log(x^4))/log(90x) = 3

I've been on it for 25 minutes and have gotten as far as:

log(x)/log(90x) = 1/2

And I'm not sure if I'm even going the right path lol, just trying to play around with properties of logs and not sure how to maneuver this. Thanks for any help
Applying the laws of logarithms to reduce the left hand sid to an exoression in $\log(x)$ :

$\dfrac{\log(x^2)+\log(x^4)}{\log(90x)}=\dfrac{2\lo g(x)+4\log(x)}{\log(90)+\log(x)}$

and you should be able to finish from there.

CB

3. Hello, DannyMath!

What you did was correct . . .

$\text{Solve for }x\!:\;\;\dfrac{\log(x^2) + \log(x^4)}{\log(90x)} \:=\: 3$

Note that: . $x \ne 0$

We have: . $\log(x^2\cdot x^4) \;=\;3\log(90x) \quad\Rightarrow\quad \log(x^6) \:=\:3\log(90x)$

. . . . . . . . . . $6\log(x) \:=\:3\log(90x) \quad\Rightarrow\quad 2\log(x) \:=\:\log(90x)$

. . . . . . . . . . . $\log(x^2) \:=\:\log(90x) \quad\;\;\Rightarrow\qquad \quad\; x^2 \:=\:90x$

Then: . $x^2 - 90x \:=\:0 \quad\Rightarrow\quad x(x-90) \:=\:0$

Therefore: . $\rlap{//////}x \,=\,0,\;\;x\,=\,90$

4. Originally Posted by Soroban
Hello, DannyMath!

What you did was correct . . .

Note that: . $x \ne 0$

We have: . $\log(x^2\cdot x^4) \;=\;3\log(90x) \quad\Rightarrow\quad \log(x^6) \:=\:3\log(90x)$

. . . . . . . . . . $6\log(x) \:=\:3\log(90x) \quad\Rightarrow\quad 2\log(x) \:=\:\log(90x)$

. . . . . . . . . . . $\log(x^2) \:=\:\log(90x) \quad\;\;\Rightarrow\qquad \quad\; x^2 \:=\:90x$

Then: . $x^2 - 90x \:=\:0 \quad\Rightarrow\quad x(x-90) \:=\:0$

Therefore: . $\rlap{//////}x \,=\,0,\;\;x\,=\,90$
UGH I was getting so frustrated and you've made it so clear. Thank you! *sigh of relief*

5. Originally Posted by DannyMath
UGH I was getting so frustrated and you've made it so clear. Thank you! *sigh of relief*
I'm afraid he has actually made rather a meal of it.

Which I suppose is OK if you are going to spoon feed someone.

CB