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