Solve for $x$ in $3^x=5^2$ by logarithm.

Question

Was solving using properties of logarithm but got stuck at the equation $x\log 3=\log5+\log5$

Answer 1

As a side note to the previous answers, $x = \frac{2\log 5}{\log 3}$ can be further simplified in the denominator to $x = \frac{\log 25}{\log 3}$, and using log rules $x = \log_3{25}$. This is a way to remove the fraction.

-FruDe

Answer 2

$$3^x=5^2\implies \log3^x=\log5^2\implies x\log3=2\log5\implies x=\frac{2\log5}{\log3}$$ I hope that helps :)

Answer 3

Hint: $\log 3$ and $\log 5$ in your equation are just constants. How would you solve for $x$ if they were replaced by say, $1$ and $2$?

Answer 4

you are almost done $$x\log3=\log5+\log5=2\log5$$ $$x=\frac{2\log5}{\log3}$$ $$x=\frac{2\cdot 0.6989}{0.4771}\approx2.93$$