Find value of $x$ in a logarithmic equation

Question

If $$2^{(\log_{2}3)^x} = 3^{( \log_3 2)^x}$$ then what is the value of $x$ in this equation? Could taking log on both sides help?

Answer 1

HINT:

With original view of the question as $2(\log_23)^x=3(\log_32)^x$

using Change of Base of Logarithm formula, $\log_23=\dfrac{\log3}{\log2}$

and take logarithm in both sides to get

$$\log2+(\log 3/\log2)^x=\log3+(\log2/\log3)^x$$

Answer 2

Let $y=(\log_2 3)^x$ and note that $(\log_{3}2)^x=\frac{1}{y}$ so you are trying to solve:

$$2^y = 3^{1/y}$$ or $$2^{y^2}=3$$

So $y=\sqrt{\log_2 3}$ and $x=1/2$. (Note: $y$ positive.)

Answer 3

note: $$\log_a b = \frac{1}{\log_b a}, \log_a b = \frac{\log b}{\log a}$$

Taking log of both sides:

$$ (\log_2 3)^x \log2 = (\log_3 2)^x \log3$$

$$ \frac{\log2}{\log3} = \frac{(\log_2 3)^{-x}}{(\log_2 3)^x}$$

$$ \log_3 2 = (\log_2 3)^{-2x} $$

$$(\log_2 3)^{-1} = (\log_2 3)^{-2x} $$

$$ x = \frac{1}{2}$$

Answer 4

Since we have $$\alpha = (\log_2 3)^x = \frac{1}{(\log_3 2)^x}$$ then your equation boils down to solving $$2^{\alpha} = 3^{1/\alpha}$$ Taking the logarithm of base $2$ on each side yields $$\alpha = \frac{1}{\alpha}\cdot \log_2 3 \iff \alpha^2 = \log_2 3$$

Back-substituting yields $$(\log_2 3)^{2x} = \log_2 3 \implies 2x = 1 \iff x = \frac{1}{2}$$