Solve $2 \log_2 5+2-x=\log_2(3^x+2^{6-x})$

Question

What I have so far: The equation can be reduced to $\frac{50}{x}=3^x+2^{6-x}$.

Since $\frac{50}{x}<0$ for $x<0$ and $3^x+2^{6-x}\geq0$ for all x there are no negative solutions.

Also since $3^3 > 50/3$ there are no solutions greater than 3 (we can maybe use this same strategy to narrow down the possibilities even more but I don think that will help much with the problem).

You can check easily that 2 is a solution and I thought that if it were true that $3^x+2^{6-x}$ is increasing for $x>2$ then since $50/x$ is decreasing it would follow that only 2 is the only solution but the former is not true.

So any other ideas ? Keep in mind that this is an algebra problem for 11 grade so advanced techniques wont help me much but feel free to share them still. Thanks.

Answer 1

It should be $\frac{100}{2^x}=3^x+2^{6-x}$. That's much easier to solve.

Answer 2

You have

$$2 \log_2 5+2-x=\log_2(3^x+2^{6-x})$$

Including the factor 2.

$$ \log_2 25+2-x=\log_2(3^x+2^{6-x})$$

Seperating $2-x$

$$ 2-x=\log_2(3^x+2^{6-x})-\log_2 25$$

$\log_c(X)-\log_c(Y)=\log_c\left(\frac{X}{Y} \right)$ so

$$2-x=\log_2\left(\frac{3^x+2^{6-x}}{25} \right)$$

Removing $\log_2()$

$$\frac4{2^x}=\frac{3^x+2^{6-x}}{25}$$

Multiplying both sides by $2^x$ and $25$

$$100=6^x+2^6$$

$$100=6^x+64$$

$$36=6^x$$

$$x=2$$