Find the value of $x$ when the powers are logarithmic functions

Question

Find the value of $x$ such that

$( \frac {x}{2} )^{\log_n 4} -(\frac {3}{x})^{\log_n 9} =0$

What I tried,

$( \frac {x}{2} )^{\log_n 4} =(\frac {3}{x})^{\log_n 9}$

Cross multiply,

$x^{\log_n 36}= (3^{\log_n 9})(2^{\log_n 4} )$

Then I noticed if I changed to base $3$ for $\log_n 9$,it would give me $\frac {2}{\log_3 n}$ or $2 \log_n 3$

Then if I changed to base $2$ for $\log_n 4$,it would give me $\frac {2}{\log_2 n}$ or $2 \log_n 2$

With different powers of both $2$ and $3$, I can't really join them up together.

Answer 1

take the logarithm of both sides we obtain $$\log_n(4)\ln(x/2)=\log_n (9)\ln(3/x)$$ using the power rule $$2\log_n(2)\ln(x/2)=2\log_n(3)\ln(3/x)$$ $$\ln(2)(\ln(x)-\ln(2))=\ln(3)(\ln(3)-\ln(x))$$ if $$\ln(n)\ne 0$$ $$\ln(2)\ln(x)-\ln(2)^2=\ln(3)^2-\ln(x)\ln(3)$$ isolating $\ln(x)$ $$\ln(x)=\frac{\ln(3)^2+\ln(2)^2}{\ln(2)+\ln(3)}$$ from here you can find $$x$$

Answer 2

Making it like $$\left( \frac {x}{2} \right)^{\alpha} -\left( \frac {3}{x}\right)^{\beta} =0$$ using your way $$\left( \frac {x}{2} \right)^{\alpha} =\left( \frac {3}{x}\right)^{\beta}\implies \frac {x^\alpha}{2^\alpha}=\frac {3^\beta}{x^\beta}$$ As you did, cross-multiply $$x^\alpha\cdot x^\beta= 2^\alpha\cdot3^\beta\implies x^{\alpha+\beta}=2^\alpha\cdot3^\beta$$ Take logarithms $$(\alpha+\beta)\log(x)=\alpha\log(2)+\beta\log(3)\implies \log(x)=\frac{\alpha\log(2)+\beta\log(3) }{\alpha+\beta }$$ and then use $x=e^{\log(x)}$.