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Comment on Powers of 3 and 5
How is -9 = -25?
Can you elaborate on your
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?
Sorry I just saw the
I'm sorry,
I guess I just don't understand the question. If x=-1, would that make the left side -9 and the right side -25? I'm just trying to understand how x=-1 so I can't move forward. Thank you for responding :)
I think I know what's
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
I did 3.3^x=5.5^x, then 3/5=5
That's a great approach. Nice
That's a great approach. Nice work!
I don't understand why x+1
We know that (3/5)^(x+1) = 1.
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
(3/5)3^x=5^x
3/5=(5/3)^x
3/5=(3/5)^-x
Then bases are equal so
-X=1
X=-1
Is it correct process?
Perfect - nice work!
Perfect - nice work!
1^0=1 and 1^1=1 too. then
Yes, 1^0 = 1 and 1^1 = 1, but
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...
So if the bases aren't equal,
I'm not entirely sure what
I'm not entirely sure what you're asking.
Once we know that (3/5)^(x+1) = 1, then we can be certain that (x+1) = 0
Does that help?
Cheers,
Brent
How can we be sure that x+1=0
We can use the fact that b^0
We can use the fact that b^0 = 1 for all non-zero values of b
For example, 6^0 = 1, 11^0 = 1, (-3)^0 = 1, (2/7)^0 = 1, (1.6)^0 = 1, etc
So, if (3/5)^(x+1) = 1, then we know that the exponent (x+1) must be zero.
Does that help?
Cheers,
Brent
On a similar note, is this
(3/5)^(1+x) = 1
(3/5)^(1+x) = (3/5)^0
1+x = 0
That's perfectly reasoned.
That's perfectly reasoned. Nice work!
I did tejureddy's way. Easy
3/5^x+1=1..after this step
Is it because we have taken the power of 1 as 0? Please let me know as I m bit confused here.
Once we know that (3/5)^(x+1)
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
I summed the 3^× to be 9^×
In your solution, you are
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
Thanks Brent. I forgot my
So, I did the following:
Convert the main problem into:
(3^1)(3^x) = (5^1)(5^x)
Turned that into:
3^x+1 = 5^x+1
At this point, I multiplied everything on the left side by 5, and everything on the right side by 3.
That left me with:
15^5x+5 = 15^3x+3
I then set 5x+5 equal to 3x+3.
Subtract 3x from both sides to get 2x+5 = 3
Subtract 5 from both sides to get 2x = -2
Divide both sides by 2, and I got x = -1.
Is this a viable way to go about this?
Hi KevinGuy,
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
it all makes sense to me
After you divide by 5^(x+1), why does (3/5)^(x+1) = 1?
I don't get where the = 1 comes from. I figured it would = 0.
Let's start from: 3^(x+1) = 5
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
Ohhhh Yes thank you very much
X can be either 0 or -1 right
x can equal -1, but it can't
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 see you explained several
I'm happy to help.
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?