# Solve this exponential equation: $3^{2x}+\left(\frac{1}{2}\right)^{-x} \cdot 3^{x+1}-2^{2x+2}=0$

I tried solving this equation $$3^{2x}+\left(\frac{1}{2}\right)^{-x} \cdot 3^{x+1}-2^{2x+2}=0$$ by taking the log of both sides, but with no results, what do I do? Sorry if this equation is very easy, I couldn't solve it...

• take $3^x =y$ i think this solves it up to some extent Commented Jul 17, 2015 at 14:21
• @Akashaero No. 
– Did
Commented Jul 17, 2015 at 14:23
• Hint: Transform this into a quadratic in $y=(3/2)^x$.
– Did
Commented Jul 17, 2015 at 14:23

## 2 Answers

Hint: This reduces to $$3^{2x} + 3\cdot 2^x \cdot 3^x - 4\cdot 2^{2x} = 0$$

If $a=3^x$ and $b=2^x$ then we get $$a^2 + 3ab - 4b^2 = 0.$$

More of a hint: You have $x = \log_3 a = \log_2 b$. So now you have two equations involving $a$ and $b$...

we have $$(3^x)^2+2^x\cdot3^x\cdot 3-(2^x)^2\cdot2^2=0$$ divided by $(2^x)^2$ we have $$\left(\frac{3^x}{2^x}\right)^2+3\cdot \frac{3^x}{2^x}-4=0$$ setting $$u=\frac{3^x}{2^x}$$ we have a quadratic equation in $$u$$

• i have got $$u_{1,2}=-\frac{3}{2}\pm\sqrt{\frac{9}{4}+\frac{16}{4}}$$ Commented Jul 17, 2015 at 15:34