# How do you solve for x in: ln(3)=ln(10)-(0.1)(x)?

### 4 Answers

Relevance

- stanschimLv 74 weeks agoFavorite Answer
Solve just like any equation:

ln(3) = ln(10) - (0.1)(x)

-0.1x = [ln(3) - ln(10)]

x = [ln(3) - ln(10)] / -0.1 = 12.03972804

- PhilipLv 64 weeks ago
ln(3) = ln(10) - (0.1)x, ie. ln(10) - ln(3) = ln(10/3) = (0.1)x. Multiply through

bt 10 getting x = 10ln(10/3) = 10(1.203972804) = 12.0397804.

- Login to reply the answers

- VamanLv 74 weeks ago
ln(3)=ln(10)-(0.1)(x)?

0.1x = ln 10 -ln 3=ln 10/3

10/3= e^(0.1x) =e^(x/10)

x = 10 ln(10/3)=12.04

- Login to reply the answers

- ?Lv 74 weeks ago
ln(3) = ln(10) - (0.1)(x) ⇒

0.1x = ln 10 - ln 3

.......ln 10 - ln 3

x = ---------------

............0.1

......2.30259 - 1.09861

x = ------------------------

............0.1

x = 12.0398......................ANS

- Login to reply the answers

Still have questions? Get your answers by asking now.

I simply subtracted ln(10) from each side and wrote the equation with the terms reversed from the original problem. Remember, try to isolate the unknown, x, on the left side of the equation when doing algebra.