# $(4x)^{\log_{10}5}=(5x)^{\log_{10}7}$

Given that $$(4x)^{\log_{10}5}=(5x)^{\log_{10}7}$$

Find $x$.

I've been trying many ways to solve it, but it ended up not correct. Hope someone can give some hints on it. Thanks in advance.

Without using logarithms you can use this approach: $$(4x)^{\log_{10}5}=(5x)^{\log_{10}7} \Rightarrow \frac{4^{\log_{10}5}}{5^{\log_{10}7}}=x^{\log_{10}7-\log_{10}5} \Rightarrow$$ $$x=\left( \frac{4^{\log_{10}5}}{5^{\log_{10}7}}\right)^{\frac{1}{\log_{10}7-\log_{10}5}}$$
taking the logarithm on both sides we get $$\log_{10} 5(\ln(4)+\ln(x))=\log_{10} 7(\ln(5)+\ln(x))$$ now set $$t=\ln(x)$$ and solve a linear equation
The equation is equivalent to $$10^{(\log_{10}(4x))(\log_{10}5)}=10^{(\log_{10}(5x))(\log_{10}7)}$$ i.e. \begin{align*} [\log_{10}(4x)]\log_{10}5&=[\log_{10}(5x)]\log_{10}7\\[3pt] (\log_{10}4+\log_{10}x)\log_{10}5&=(\log_{10}5+\log_{10}x)\log_{10}7\\[3pt] (\log_{10}5-\log_{10}7)\log_{10}x&=(\log_{10}7-\log_{10}4)\log_{10}5\\[3pt] \log_{10}(\tfrac57)\log_{10}x&=\log_{10}(\tfrac74)\log_{10}5\\[3pt] \log_{10}x&=\frac{\log_{10}(\frac74)\log_{10}5}{\log_{10}(\frac57)} \end{align*} Then $$x=10^{\frac{\log_{10}(\frac74)\log_{10}5}{\log_{10}(\frac57)}}=5^{\log_{5/7}(7/4)}$$