Let $\log_{10}2 = x$ and $\log_{10}3 = y$. Express the $\log_{10}5$ in terms of $x$ and/or $y$.
I am stuck on finding a way to express $5$ as a product or quotient of $x$ and $y$.
Any help, comment, or suggestions on how to solve this problem will be much appreciated. Thank you in advance.
The simplest answer is $1-x$ $$1-x$$ $$\log_{10}(10)-\log_{10}(2)$$ $$\log_{10}(\frac{10}{2})$$ $$\log_{10}(5)$$
Reversing some of the given log's almost always worked for me: $$\lg5=A \implies \color{red}{\log_52}+1=\frac{1}{A}$$ $$x=\lg2 \implies \log_25+1=\frac{1}{x} \implies \log_52=\frac{1}{\frac{1}{x}-1}=\frac{x}{1-x}$$ $$\frac{1}{A}=\color{red}{\frac{x}{1-x}}+1=\frac{1}{1-x}\implies \boxed{A=1-x}$$
