Derive algebraic solution for $x^{\log_3 2} = \sqrt x + 1$

Question

I want algebraic solution to $$x^{\log_3 2} = \sqrt x + 1.$$ Computer solution says $x = 9$, but I've failed in several attempts to derive this answer algebraically. Any help is much appreciated.

Answer 1

Note that $$a^{\log_b c}=c^{\log_b a}$$

Sothe equation becomes $$2^{\log_3 x}=\sqrt{x}+1$$ You may only verify that $x=9$ is a roots as $$2^{\log_3 9}=2^2=4$$ and $$4=\sqrt{9}+1.$$

Answer 2

Below is an algebraic derivation. Reexpress both sides of the equation,

$$x^{\log_3 2} = \sqrt x + 1$$

as

$$RHS = \sqrt x + 1 = 3^{\log_3 \sqrt x} + 1 = 3^{\frac 12 \log_3 x}+ 1^{\frac 12 \log_3 x}$$

$$LHS = x^{\log_3 2} = (3^{log_3 x})^{\log_3 2}=(3^{log_3 2})^{\log_3 x}=2^{\log_3 x} =4^{\frac 12 \log_3 x}$$

As a result, the original equation can be expressed in a form of identical exponents,

$$4^{\frac 12 \log_3 x} = 3^{\frac 12 \log_3 x}+ 1^{\frac 12 \log_3 x}$$

Since $4=3+1$, the exponent must be one, i.e.

$$\frac 12 \log_3 x=1$$

which leads to the solution,

$$x=9$$