Solve for x when $(\log_x (5x))(\log_7 x)=2$

Question

I've been trying to use the change of base property but I'm not having much luck. Can anyone give me any ideas on how I should approach this problem? The answer is 49/5

Thanks.

Answer 1

Using change-of-bases, we have $\log_x(5x) \cdot \log_7(x)= \frac{\log (5x)}{\log (x)} \cdot \frac{\log (x)}{\log (7)} = \frac{\log (5x)}{\log 7} = \log_7(5x)= 2$, where the base of the intermediate $\log$s do not really matter.

Answer 2

Hint: $$\log_x(5x)=\frac{2}{\log_7 x} = \frac{\log_7 49}{\log_7 x}$$ I hope that you have tested the conditions for the logarithms to be defined.

Answer 3

Hint: Use properties of logs: $\log xy=\log x+\log y\\\log\frac xy=\log x-\log y\\\log a^x=x\log a\\b^{\log_bx}=x=\log_bb^x\\\log_ba=\frac{\log_xa}{\log_xb}$.

Answer 4

$$\log_x 5x \cdot \log_7 x = 2$$ $$\log_x 5x = \frac{2}{\log_7 x}$$ $$\log_x 5x = \frac{2\log_x 7}{log_x x}$$ $$\log_x 5x = 2\log_x 7$$ $$\log_x 5x = \log_x 49$$ $$\implies 5x = 49$$ $$x =\frac{49}{5}$$