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

Question

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

So you have $x(a)=2b$ and you have to solve for $x$. – Anurag A Jul 14 '20 at 14:12 — Anurag A, Jul 14 '20 at 14:12
If that's a minus sign on left need to adjust solution. – coffeemath Jul 14 '20 at 14:15 — coffeemath, Jul 14 '20 at 14:15
Ask yourself,how would you solve for $x$ in $5x=7$? – Anurag A Jul 14 '20 at 14:15 — Anurag A, Jul 14 '20 at 14:15
We can simply write $x=\log_3(5^2)$ :) – K.defaoite Jul 14 '20 at 14:21 — K.defaoite, Jul 14 '20 at 14:21

score 2 · Answer 1 · answered Jul 14 '20 at 14:17

2

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 $2$?

answered Jul 14 '20 at 14:17

paulinho

6,553

Thanks a lot paulino for helping – Maths Jul 14 '20 at 14:21

TShiong · Answer 2 · 2020-07-14T14:23:37.533

0

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

edited Jul 14 '20 at 14:23

answered Jul 14 '20 at 14:18

TShiong

1,257

1

Thanks for helping – Maths Jul 14 '20 at 14:22

score 0 · Accepted Answer · answered Jul 14 '20 at 17:49

0

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

answered Jul 14 '20 at 17:49

Sirswagger21

1,841

Thank u for sol – Maths Jul 14 '20 at 18:09
Sure thing @Maths – Sirswagger21 Jul 14 '20 at 18:16

score -1 · Answer 4 · answered Jul 14 '20 at 14:17

-1

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

answered Jul 14 '20 at 14:17

A-Level Student

6,520

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

4 Answers4