# Solve the equation $2^{2x}+2^{2x-1}=3^{x+0.5}+3^{x-0.5}$

Solve the equation $$2^{2x}+2^{2x-1}=3^{x+0.5}+3^{x-0.5}$$

The given equation is equivalent to $$2^{2x}+\dfrac12\cdot2^{2x}=3^x\sqrt3+\dfrac{3^x}{\sqrt3}$$ which is $$\dfrac32\cdot2^{2x}=3^x\left(\sqrt3+\dfrac{1}{\sqrt3}\right)$$ The last equation can be written as $$\dfrac32\cdot2^{2x}=\dfrac{4}{\sqrt{3}}\cdot3^x,$$ or $$\dfrac{2^{2x}}{3^x}=\dfrac{\frac{4}{\sqrt3}}{\frac{3}{2}}\iff\left(\dfrac43\right)^x=\dfrac{\frac{4}{\sqrt3}}{\frac{3}{2}}$$ How do I find $$x$$ from here, as it is obviously a difficulty for me? What's the approach supposed to be?

• Note that both sides are monotone increasing in $x$, with the $4^x$ side dominating eventually. There is only one intersection which can be found by inspection (with a little effort).
– lulu
Commented Jan 15, 2023 at 15:05
• I haven't checked your algebra. If it's right then all you need at the last step is $\log_{4/3}$. Commented Jan 15, 2023 at 15:06
• $\dfrac{\frac{4}{\sqrt3}}{\frac{3}{2}}=\frac{4}{3}\frac{2}{\sqrt{3}}=\frac{4}{3}\sqrt{\frac{4}{3}}=(\frac{4}{3})^{\frac{3}{2}}$
– Sil
Commented Jan 15, 2023 at 15:11
• take $x = t + \frac{1}{2}$ to remove squared roots...comes out $\left( \frac{4}{3} \right)^t = \frac{4}{3}$ Commented Jan 15, 2023 at 15:40

Alternative approach:

• Let $$r = \log_2(3).$$

• Note that $$2^a = 2^b \iff a = b.$$

Solve the equation $$2^{2x}+2^{2x-1}=3^{x+0.5}+3^{x-0.5}$$

The LHS can be re-written as

$$2^{2x} \times \left[1 + \frac{1}{2}\right] = 2^{2x} \times \frac{3}{2}.$$

The RHS can be re-written as

$$3^{x + 0.5} \times \left[1 + \frac{1}{3}\right] = 3^{x + 0.5} \times \frac{4}{3}$$

$$= \left(2^{r}\right)^{x+0.5} \times \frac{4}{3} = \left[2^{r \times (x+0.5)}\right] \times \frac{4}{3}.$$

Since the LHS and RHS are equivalent, you know that

$$2^{2x-3} = \frac{2^{2x}}{8} = \frac{\left[2^{r \times (x+0.5)}\right]}{9} = 2^{r \times (x + 0.5 -2)}.$$

Therefore,

$$2^{2x-3} = 2^{r \times \left[x - \frac{3}{2}\right]} \implies$$

$$2x - 3 = r \times \left[x - \frac{3}{2}\right] \implies$$

$$x(2 - r) = 3 - \left[r \times \frac{3}{2}\right] \implies$$

$$x = \frac{3 - \left[r \times \frac{3}{2}\right]}{2 - r} = \frac{3}{2}.$$

Since your last step then $$(4/3)^{x}=(4/3)^{3/2}$$ then $$x=3/2$$ as Sil noticed.

Alternatively, let $$x$$ be a real number, then

\begin{align*} 2^{2x}+2^{2x-1}&=3^{x+1/2}+3^{x-1/2}\\ \frac{1}{3^{x+1/2}}\left(2^{2x}+2^{2x-1} \right)&=\frac{1}{3^{x+1/2}}\left(3^{x+1/2}+3^{x-1/2} \right)\\ 2^{2x}\cdot3^{-(x+1/2)}+2^{2x-1}\cdot 3^{-(x+1/2)}&=3^{x+1/2}\cdot 3^{-(x+1/2)}+3^{x-1/2}\cdot 3^{-(x+1/2)}\\ 2^{2x}\cdot3^{-(x+1/2)}+2^{2x-1}\cdot 3^{-(x+1/2)}&=4/3\\ \underbrace{e^{\ln 2^{2x}}\cdot e^{\ln 3^{-(x+1/2)}}}_{(*)}+e^{\ln 2^{2x-1}}\cdot e^{\ln 3^{-(x+1/2)}}&=4/3 \end{align*}

For $$(*)$$ note that is just $$e^{(2x)\ln 2-(x+1/2)\ln 3}:=t$$, then the above equation came be written as $$\frac{3t}{2}=\frac{4}{3}$$, that is, $$t=8/9$$.

Now, we need to find $$x$$ into $$8/9=e^{(2x)\ln 2-(x+1/2)\ln 3}$$. Applying logarithm both sides we have $$(2x)\ln 2-(x+1/2)\ln 3=\ln(8/9)$$. Rewriting in order to find $$x$$ we have $$(2\ln 2-\ln 3)x=1/2\ln 3+\ln(8/9)$$. Thus ,

$$\boxed{x}=\frac{\frac{\ln 3}{2}+\ln\frac{8}{9}}{2\ln 2-\ln 3}=\boxed{\frac{\frac{1}{2}\ln 3+3\ln 2-2\ln 3 }{2\ln 2 -\ln 3}=\frac{3}{2}}$$