Solving Exponential Equations with Equal Bases?

Question

1. 9^2x = 27^x - 1

2. 5^n-1 = 1/25

3. 25^x = 5^x2 - 15

4. 2^x * 4^x+5 = 4^2x-1

5. (sqrt3)^2x+4 = 9^x-2

Answer 1

Hmmm. Once the bases are equal, you can ignore them, and just go about setting the exponents equal.

1. 9^2x = 27^x - 1

(3^2)^2x = (3^3)^x-1
3^(4x) = 3^(3x-3)
4x = 3x - 3

etc.

Is that enough?

Answer 2

Okay, the key thing in these problems is to get them all to become the same base.

1. 9^2x = 27^x-1

Let's look at that 27 and 9.... What base can we get to make life easier here?

9 = 3^2
27 = 3^3

Great! 3^2 and 3^3 have the same base! Watch this next step.

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

Because of the same base, you can do it just as if it were an equation. Minus the base... So you'll get...

2(2x) = 3(x-1)
4x = 3x - 3
x = -3

There's your answer.

2. 5^n-1 = 1/25

Now the fraction problems are the uh-oh's, and the last one I'll go with, because it's 1:12 AM where I'm at but anyway, remember that if...

5^1 = 5

Then it's inverse will be

5^-1 = 1/5

It turns out to be a fraction! So... You can turn that 1/25 into something neat... Maybe raise it to the -2 Power? That will do! Now watch what happens when you do that

5^n-1 = 5^-2

SAME BASES! AGAIN! AH HAHA! Now, do what you do best and end up with...

n-1 = -2
n = -1

Because I stayed up late, I could be wrong. Someone can check this over.

Answer 3

Only one problem per question please.

Please learn to use parentheses.

5. (sqrt3)^2x+4 = 9^x-2

I assume you mean

(√3)^(2x + 4) = 9^(x - 2)
[3^(1/2)]^(2x + 4) = (3^2)^(x - 2)
3^(x + 2) = 3^(2x - 4)

x + 2 = 2x - 4
x = 6