# Adding logarithms with different bases

Just had an exam, this sinister question I know I did wrong lingers in my mind: Solve for $x$,

$$2-\log_3(x-7) = \log_{\frac{1}{3}} (2x)$$

On phone not sure how to write the equation properly.

Please correct and take me through solving the problem. (I have little problems with same base logarithmic problems)

Thanks

• Hint: Use $\log_b x = \frac{\ln x}{\ln b}$. – gammatester Apr 15 '16 at 7:36
• Hint: $\frac{1}{3}^x=3^{-x}$, so you can use that $\log_{1/3} x = - \log_3 x$. – Édes István Gergely Apr 15 '16 at 7:37

$$2-\log_3(x-7)=\log_{\frac13}(2x)$$ $$2-\frac{\log(x-7)}{\log3}=\frac{\log(2x)}{\log{\frac13}}$$ $$2-\frac{\log(x-7)}{\log3}=\frac{\log(2x)}{-\log{3}}$$ $$2\log(3)-\log(x-7)=-\log(2x)$$ $$\log{\frac{9}{x-7}}=\log{\frac{1}{2x}}$$ $$\frac{9}{x-7}=\frac{1}{2x}$$ $$18x=x-7$$ $$x=\frac{-7}{17}$$ But for logarithm to be defined, we must have $x-7\gt0$ and $2x\gt0$. Thus, there is no solution.