Question

How do I solve for x in the equation 5^x=3125? Also in 5^x=1/125, 9^x=243, 4^x=1/32, and 3*2^x=96?

Answer 1

Remember
if x^m=x^n and x=x, m=n
also
x^-m=1/(x^m)
(x^m)^n=x^(mn)

try to match bases

5^x=3125
5^x=5^5
x=5

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


9^x=234
3^(2x)=3^5
2x=5
x=2.5


4^x=1/32
1/32=1/(2^5)=2^-5
2^(2x)=2^-5
2x=-5
x=-2.5


3*2^x=96
first divide both sides by 3
2^x=32
2^x=2^5
x=5




5^x=3125
x=5

5^x=1/125
x=-3


9^x=234
x=2.5

4^x=1/32
x=-2.5

3*2^x=96
x=5

Answer 2

5^x=1/125

Take log of both sides

log(5^x)=log(1/125)

x*(log(5))= log(1/125)

x= log(1/125)/ log(25)

x=-3

9^x= 243

Take log of both sides

log(9^x)= log(243)

x*(log(9))= log(243)

x= log(243)/ log(9)

x= 2.5

4^x= 1/32

Take log of

log(4^x)=log(1/32)

x*(log(4))=log(1/32)

x= log(1/32)/log(4)

x= -2.5

3*2^x=96

Divide both sides by 3 so we can make it more simply

3(2^x)/3= 96/3

2^x= 32

Take log of

log(2^X)=log(32)

x(log(2))= log(32)

x= log(32)/log(2)

x=5

As you can see, I only used a simple way for all of them.

I hope that's help:0