Thread: 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.

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

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?

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

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



Remeber that



Now cross multiply bth sides



Take the cube root on both sides and you get

Ahh, thanks!