# Thread: Two logarithmetic problems, solve for x

1. ## Two logarithmetic problems, solve for x

Attached are the problems.

For the first problem, the x cancels out each other. So how do I solve for x?

For the second problem, I don't know how to make the x^-3 into a positive exponent so I can cube root both sides.

2. Portion 1

(x+2)log5 = (3+x)log9
x+2/3+x = log9/log5
x+2/3+x = 1.37 ( cross multiply )
x+2 = 4.11 + 1.37x
-0.37x = 2.11
x = -5.7

2nd portion

x^-3 = 125
1 / x^3 = 125
x^3 = 1/125
x = Cube root(1/125)
x = 1/5

3. How did you do the second portion? Like what do I do when there is a negative exponent and I have to take the root of both sides?

4. x^-3 = 125
1 / x^3 = 125
x^3 = 1/125
x = Cube root(1/125)
x = 1/5

5. Originally Posted by mwok
For the second problem, I don't know how to make the x^-3 into a positive exponent so I can cube root both sides.

Well, this is how you do the second part

$\displaystyle x^{-3} = 125$

Remeber that $\displaystyle x^{-3}= \frac{1}{x^{3}}$

$\displaystyle \frac{1}{x^{3}} = 125$

Now cross multiply bth sides

$\displaystyle x^{3} = \frac{1}{125}$

Take the cube root on both sides and you get

$\displaystyle x = \frac{1}{5}$

6. Ahh, thanks!