Question: Powers of 3 and 5

I'm pretty sure we're not suggesting that -9 = -25 :-)
The solution to the equation is x = -1
If we replace x with -1 at the point in our solution where we have 3^(x+1) = 5^(x+1), we get 3^0 = 5^0, which is the same as saying 1 = 1.
Does that help?

I think I know what's happening.
3^(-1) = 1/3 and 5^(-1) = 1/5
So, 3^x + 3^x + 3^x = 5^x + 5^x + 5^x + 5^x + 5^x evaluates to be:
1/3 + 1/3 + 1/3 = 1/5 + 1/5 + 1/5 + 1/5 + 1/5

I think you're evaluating 3^(-1) as -3
For more, see the following video:
- Negative exponents: https://www.greenlighttestprep.com/module/gre-powers-and-roots/video/1029

That's a great approach. Nice work!

We know that (3/5)^(x+1) = 1.
If (3/5)^(something) = 1, then something must equal 0
There is no other exponent such that (3/5)^exponent = 1

Perfect - nice work!

Yes, 1^0 = 1 and 1^1 = 1, but that does not affect the solution to the given equation, since the bases are 3 and 5 (not 1).

If the bases are equal (e.g., 3^(x - 2) = 3^5x), then we can conclude (usually) that the exponents are equal. However, in your simplification (3/5)^x+1 = 1^1, the bases are NOT equal.

More here: https://www.greenlighttestprep.com/module/gre-algebra-and-equation-solvi...

Once we know that (3/5)^(x+1) = 1, we can ask:
If (3/5)^SOMETHING = 1, what can we conclude about SOMETHING?
Well, we know that (3/5)^0 = 1
So, it must be the case that SOMETHING = 0

In other words, it must be the case that (x+1) = 0

Does that help?

Cheers,
Brent

In your solution, you are adding the bases to get 3^x + 3^x + 3^x = 9^x
Unfortunately, there is no such property when it comes to adding powers.

We can quickly test your hypothesis by assigning a value to x and checking whether your simplification works.
Let's say x = 2
Is it the case that 3^2 + 3^2 + 3^2 = 9^2?
Evaluate each side to get: 9 + 9 + 9 = 81. This is not true.
So, we can't say that 3^x + 3^x + 3^x is equal to 9^x

Instead, we must use a certain rule when it comes to simplifying expressions.
Notice that:
10 + 10 + 10 = 3(10)
2 + 2 + 2 = 3(2)
x + x + x = 3x
k² + k² + k² = 3k²
5n + 5n + 5n = 3(5n)
In general, (something) + (something) + (something) = 3(something)

Likewise, 3^x + 3^x + 3^x = 3(3^x) = (3^1)(3^x) = 3^(x+1)
Does that help?

Cheers,
Brent

Hi KevinGuy,

Sorry for the delay. I don't know how this question eluded me earlier.

Unfortunately, it's just a coincidence that you arrived at the correct answer.

Your solution is fine up to the point where you got: 3^(x+1) = 5^(x+1)

When you got to that point, you multiplied everything on the left side by 5, and everything on the right side by 3.
This breaks the golden rule of equation solving: what you do to one side of the equation, you must do the SAME THING to the other side.

Consider this example equation: 2x = 6
If we multiply the left side by 5, and right side by 3, we get: 10x = 18.
In the first equation, the solution is x = 3
In the second equation, the solution is x = 1.8

Cheers,
Brent

Let's start from: 3^(x+1) = 5^(x+1)

When we divide BOTH sides by 5^(x+1), we get: 3^(x+1)/5^(x+1) = 5^(x+1)/5^(x+1)
In the fraction 5^(x+1)/5^(x+1), the numerator and denominator are the same.
This means 5^(x+1)/5^(x+1) = 1

Does that help?

Cheers,
Brent

x can equal -1, but it can't equal 0.

Here's why:
Take the given equation: 3^x + 3^x + 3^x = 5^x + 5^x + 5^x + 5^x + 5^x
Replace x with 0 to get: 3^0 + 3^0 + 3^0 = 5^0 + 5^0 + 5^0 + 5^0 + 5^0
Evaluate: 1 + 1 + 1 = 1 + 1 + 1 + 1 + 1
We get: 3 = 5, which is incorrect.
So, x = 0 can't be a solution.

I'm happy to help.

Let's first review a few key concepts:

CONCEPT #1: k^0 = 1 for all non-zero values of k
For example, 6^0 = 1, 17^0 = 1, (4/3)^0 = 1. etc

CONCEPT #2: If x^a = x^b, then a = b (as long as x ≠ 0, x ≠ 1, x ≠ -1)
For example, if 9^x = 9^7, then x = 7
Likewise, if (11.8)^x = (11.8)^54, then x = 54

Now for the question...

Let's start from here: (3/5)^(x+1) = 1
Rewrite the right-hand side as follows: (3/5)^(x+1) = (3/5)^0 [this works because of CONCEPT #1]
Apply CONCEPT #2 to get:(x+1) = 0

Does that help?