I'm currently working on a math problem that involves solving an exponential equation with fractions. Here's the equation:
$$ \frac{3^x+2^x}{3^x-2^x}=7 $$
I've been trying to solve it on my own, but I'm stuck and would really appreciate some help. My attempt at solving the problem involved multiplying both sides by the denominator, which gave me:
$$ 3^x+2^x=7(3^x-2^x) $$
Expanding the right-hand side, I got:
$$ 3^x+2^x=7\cdot 3^x-7\cdot 2^x $$
Then I moved all the terms involving $3^x$ to the left-hand side and all the terms involving $2^x$ to the right-hand side, which gave me:
$$ 9\cdot 2^x=6\cdot 3^x $$
Dividing both sides by $2^x\cdot 3^x$, I got:
$$ \frac{9}{6}=\left(\frac{3}{2}\right)^x $$
Simplifying, I got:
$$ \left(\frac{3}{2}\right)^x=\frac{3}{2} $$
And that's where I got stuck. I'm not sure how to proceed from here to find the largest solution to the equation. Can anyone give me a hint or point me in the right direction?
Thank you in advance for your help!
Your work is fine, as noticed you have made a mistake here
$$3^x+2^x=7\cdot 3^x-7\cdot 2^x \implies 8\cdot 2^x=6\cdot 3^x \implies \left(\frac{3}{2}\right)^x=\frac{4}{3}$$
from which we can easily conclude using logarithm.
As an alternative manipulation, using that
$$\frac{3^x+2^x}{3^x-2^x}=7 \iff \frac{\left(\frac{3}{2}\right)^x+1}{\left(\frac{3}{2}\right)^x-1}=7$$
and by $y=\left(\frac{3}{2}\right)^x-1$ we obtain
$$\frac{y+2}{y}=1+\frac2y=7 \implies y=\frac13 \implies \left(\frac{3}{2}\right)^x=\frac{4}{3}$$