How to solve for x? 25^(3x-1) = 125^(x+2)?

Question

25^(3x-1) = 125^(x+2)

It says not to use calculators. I know I've done this type of problem before, but I totally forgot how. For some reason, I'm thinking I'm supposed to use log? Help!

Answer 1

25^(3x - 1) = 125^(x + 2)

5^[2(3x - 1)] = 5^[3(x + 2)]

2(3x - 1) = 3(x + 2)

6x - 2 = 3x + 6

6x - 3x = 2 + 6

3x = 8

x = 8/3

Answer 2

25^(3x-1) is 5^(2*(3x-1))

125^(x+2)=5^(3*(x+2)

So you have

5^(6x-2)=5^(3x+6)

The bases are the same. Set the exponents equal

6x-2=3x+6

3x=8

x=8/3

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Answer 3

No you dont need to use logs but i can understand why you were thinking that. Because log questions use the same technique that needs to be used in this question

First of all you need to make the bases same

5^2 = 25

5^3 = 125

5^2(3x - 1) = 5^3(x+2)

Now since base is the same now, you can bring down the exponents

2 (3x - 1) = 3 (x + 2)

6x - 2 = 3x + 6

3x = 8

x = 8/3

You do similar thing in logs i.e. making bases equal and bringing down exponents but if you had to use logs it would have been in the question :)

Hope this helps :)