Solve in terms of $ln(5)$ and $ln(3)$: $15^{4y}=3^{4y+8}$

Question

Here is my attempt, and I don't understand why it doesn't work (I apologise for my clunky MathJax):

$(4y)ln15=(4y+8)ln3$

$(4y)(ln3+ln5)=(4y+8)ln3$

$4(y(ln3+ln5)=4(y+2ln3)$

$y(ln3+ln5)=(y+2ln3)$

$y(ln3+ln5)-y=(2ln3)$

$y(ln3+ln5-1)=(2ln3)$

$y=\frac{2ln3}{ln3+ln5-1}$

The correct answer, however, is $\frac{2ln3}{ln5}$. Could someone point to my mistake, as I can't recognise it myself? Thank you.

Answer 1

Here is the mistake:

$$4y(\ln3+\ln5)=4(y+2\color{red})\ln3$$

$$y(\ln 3 + \ln 5) = (y+2) \ln 3$$

$$y\ln 5 = 2\ln 3$$

Answer 2

i think it must be $$4y(\ln(3)+\ln(5))=(4y+8)\ln(3)$$

Answer 3

Mistake is in step $2$ and $3$ $$(4y+8)\ln 3\ne 4(y+2\ln3)$$