0
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

Without using logarithms you can use this approach: $$(4x)^{\log_{10}5}=(5x)^{\log_{10}7} \Rightarrow \frac{4^{\log_{10}5}}{5^{\log_{10}7}}=x^{\log_{10}7-\log_{10}5} \Rightarrow$$ $$x=\left( \frac{4^{\log_{10}5}}{5^{\log_{10}7}}\right)^{\frac{1}{\log_{10}7-\log_{10}5}}$$

$\endgroup$
3
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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 $$x=10^{\frac{\log_{10}(\frac74)\log_{10}5}{\log_{10}(\frac57)}}=5^{\log_{5/7}(7/4)}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.