# Solve for $x$ in $3^x=5^2$ by logarithm. [closed]

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

• So you have $x(a)=2b$ and you have to solve for $x$. Jul 14, 2020 at 14:12
• Yes dont know the answer Jul 14, 2020 at 14:14
• If that's a minus sign on left need to adjust solution. Jul 14, 2020 at 14:15
• Ask yourself,how would you solve for $x$ in $5x=7$? Jul 14, 2020 at 14:15
• We can simply write $x=\log_3(5^2)$ :) Jul 14, 2020 at 14:21

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 $$2$$?

• Thanks a lot paulino for helping Jul 14, 2020 at 14:21

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

• Thank u for sol Jul 14, 2020 at 18:09
• Sure thing @Maths Jul 14, 2020 at 18:16

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

• Thanks for helping Jul 14, 2020 at 14:22

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