Hard exponential equation

Question

Is there any algebraic way of solving the following equation for $x$?

$$\frac{3^x+2^x}{3^x-2^x}=7$$

Apparently there is some way of solving this and I heve tried to solve it in a conventional algebraic method without success.

The answer should be:

$$\frac{2\ln(2)-\ln(3)}{\ln(3)-\ln(2)}$$

Thanks in advance

Answer 1

$$3^x+2^x=7\times3^x-7\times2^x\iff 7\times2^x+2^x=7\times 3^x-3^x$$ $$\iff 8\times 2^x=6\times 3^x$$ Can you finish it?

Answer 2

Hint: Write $$3^x+2^x=7(3^x-2^x)$$ so we get $$\left(\frac{3}{2}\right)^x+1=7\left(\left(\frac{3}{2}\right)^x-1\right)$$ now let $$\left(\frac{3}{2}\right)^x=t$$ $$t+1=7(t-1)$$

Answer 3

Yes. Multiply through by $3^x-2^x$ to get $$ 3^x+2^x=7(3^x-2^x)=7(3^x)-7(2^x) $$ or equivalently $$ 8(2^x)=6(3^x) $$ i.e. $$ 2^{x+3}=2(3^{x+1}) $$ or $$ 2^{x+2}=3^{x+1} $$ which you can hopefully solve by taking logs.

Answer 4

multiply the denominator $$3^x+2^x = 7(3^x-2^x)$$ distribute $$3^x+2^x = 7\cdot 3^x-7 \cdot 2^x$$ sum $-7 \cdot 3^x - 2^x$ to both sides $$-6 \cdot 3^x = -8 \cdot 2^x$$ which simplifies to $$\left( \frac{3}{2} \right)^x=\frac{8}{6}= \frac{4}{3}$$ so the answer is $$x= \frac{\ln (4/3)}{\ln (3/2)} = \frac{2 \ln 2 - \ln 3}{\ln 3 - \ln2}$$

Answer 5

$$3^x+2^x=7(3^x)-7(2^x)$$ $$2^x(8)=3^x(6)$$ $$(\frac{3}{2})^x=\frac{4}{3}$$ take $\log$ $$x\log(\frac{3}{2})=\log\frac{4}{3}$$