Exponential equation

[Kalen12]

9*16^x + 4*81^x = 13*36^x

The goal here is to find the sum of all real solutions to this equation. But only the x found for me will suffice. The given answers are:
a)-1 b)1/2 c)1 d)-1/2

Best regards

 

[Deleted member 4993]

9*16^x + 4*81^x = 13*36^x

The goal here is to find the sum of all real solutions to this equation. But only the x found for me will suffice. The given answers are:
a)-1 b)1/2 c)1 d)-1/2

Best regards

Please refer to our guidelines of posting in this forum enunciated at: https://www.freemathhelp.com/forum/threads/guidelines-summary.109845/

Notice that the given expression can be written in terms of "power" 2 & 3 (except of course 13)

Please share your work/thoughts with us.

 

[Kalen12]

So here is the question again since I can't find the edit button.

9×16^x+4×81^x=13×36^x

I'm stuck at this exponential equation while practicimg for a test. The only work I managed to done with it so far is:

3²×4^2x+2²×9^2x=13×6^2x

I tried to solve it in several ways but it just turns out illogical or I do stuff which I nornally can't in an equation. One of the things I tried is switching 2x with t 2x=t and trying something by that but even with that I don't have a clue of how to proceed.
Any help is welcome

Best regards

 

[Deleted member 4993]

9*16^x + 4*81^x = 13*36^x

The goal here is to find the sum of all real solutions to this equation. But only the x found for me will suffice. The given answers are:
a)-1 b)1/2 c)1 d)-1/2

Best regards

9*16^x + 4*81^x = 13*36^x

9*4*4^2x-1 + 4*9*9^2x-1 = 13*36*36^x-1

4^2x-1 + 9^2x-1 = 13*36^x-1

continue....

 

[Kalen12]

Firstly how did you get that -1 in the exponent? I'm guessing that you can add it just like you can, for example, square the whole equation or multiply it by (-1), right?

2^4x-2+3^4x-2=13×6^4x-2

 

[Deleted member 4993]

Firstly how did you get that -1 in the exponent? I'm guessing that you can add it just like you can, for example, square the whole equation or multiply it by (-1), right?

2^4x-2+3^4x-2=13×6^4x-2

No! That should be-

2^4x-2+3^4x-2=13×6^2x-2

 

[Kalen12]

So then how did you get that -1. I am confused.
Okay so what from that point, still don't see a way.

 

[Deleted member 4993]

So then how did you get that -1. I am confused.
Okay so what from that point, still don't see a way.

Where are you seeing that I got "-1"?

The answer is "1" or "+1".

 

[ksdhart2]

So then how did you get that -1. I am confused.
Okay so what from that point, still don't see a way.

Consider what exponentiation really means - it's repeated multiplication. When given an expression like \(36^{x}\), we can also write that as:

\(36 \cdot 36 \cdot 36 \cdot 36 \cdots 36 \cdot 36\)

Where there are \(x\) copies of 36. Now, since multiplication is commutative and associative, we can freely insert some parentheses:

\(36^x = 36 \cdot \left( 36 \cdot 36 \cdots 36 \cdot 36 \right) \)

How many copies of 36 are inside the parentheses? Well, we had \(x\) before and we left one outside, so there must be \(x - 1\) inside. Hence:

\(36^x = 36 \cdot 36^{x - 1} \)

As for how to proceed further, let's look for a moment only at the right-hand side of the equation:

\(13 \cdot 6^{2x-2}\)

Can you see why we can also write this as:

\(13 \cdot \left(2 \cdot 3\right)^{2x-2}\)

Can you think of a rule of exponentiation that lets you do something with this? Perhaps one on the page I linked above? Bear in mind that we've got a power of 2 and a power of 3 on the left-hand side. It sure would be great if we could get a power of 2 and a power of 3 on the right-hand side too, maybe then make use of a different rule of exponentiation...

 

[MarkFL]

To make things easier, I would use the following substitutions:

[MATH]u=4^x[/MATH]
[MATH]v=9^x[/MATH]
And then the equation becomes, when suitably arranged:

[MATH]9u^2-13uv+4v^2=0[/MATH]
Factor:

[MATH](9u-4v)(u-v)=0[/MATH]
Case 1:

[MATH]9u-4v=0[/MATH]
[MATH]4^{x-1}=9^{x-1}[/MATH]
Case 2:

[MATH]u-v=0[/MATH]
[MATH]4^{x}=9^{x}[/MATH]
In both cases, the equations are only true when the exponents are zero, meaning:

[MATH]x\in\{0,1\}[/MATH]

 

[Kalen12]

Where are you seeing that I got "-1"?

The answer is "1" or "+1".

Sorry, I meant that "-1" in the exponent (4^2x-1)

 

[Kalen12]

To make things easier, I would use the following substitutions:

[MATH]u=4^x[/MATH]
[MATH]v=9^x[/MATH]
And then the equation becomes, when suitably arranged:

[MATH]9u^2-13uv+4v^2=0[/MATH]
Factor:

[MATH](9u-4v)(u-v)=0[/MATH]
Case 1:

[MATH]9u-4v=0[/MATH]
[MATH]4^{x-1}=9^{x-1}[/MATH]
Case 2:

[MATH]u-v=0[/MATH]
[MATH]4^{x}=9^{x}[/MATH]
In both cases, the equations are only true when the exponents are zero, meaning:

[MATH]x\in\{0,1\}[/MATH]

Thanks this helps a lot, but I can't help but notice that you got 4^x-1 and 9^x-1 in the first case. Now I'm still wondering how that -1 ended up there?

When I take this post into consideration and try to solve this equation I get this:

Step-by-step:

9*16^x+4*81^x=13*36^x

9*(4^x)²+4*(9^x)²=13*(4*9)^x (or just 13*4^x*9^x)

4^x=u
9^x=v

9u²+4v²=13uv

9u²-13uv+4v²=0

(9u-4v)(u-v)=0

Case 1:
9u-4v=0
9*4^x=4*9^x



Case 2:
u-v=0
u=v
4^x=9^x




And from this point I'm not sure what to do since I am a bit rusty with logarithms, but I know that we can solve both these cases with logarithms and get the values you mentioned earlier.

 

[Kalen12]

Consider what exponentiation really means - it's repeated multiplication. When given an expression like \(36^{x}\), we can also write that as:

\(36 \cdot 36 \cdot 36 \cdot 36 \cdots 36 \cdot 36\)

Where there are \(x\) copies of 36. Now, since multiplication is commutative and associative, we can freely insert some parentheses:

\(36^x = 36 \cdot \left( 36 \cdot 36 \cdots 36 \cdot 36 \right) \)

How many copies of 36 are inside the parentheses? Well, we had \(x\) before and we left one outside, so there must be \(x - 1\) inside. Hence:

\(36^x = 36 \cdot 36^{x - 1} \)

As for how to proceed further, let's look for a moment only at the right-hand side of the equation:

\(13 \cdot 6^{2x-2}\)

Can you see why we can also write this as:

\(13 \cdot \left(2 \cdot 3\right)^{2x-2}\)

Can you think of a rule of exponentiation that lets you do something with this? Perhaps one on the page I linked above? Bear in mind that we've got a power of 2 and a power of 3 on the left-hand side. It sure would be great if we could get a power of 2 and a power of 3 on the right-hand side too, maybe then make use of a different rule of exponentiation...

Well this makes things a bit clearer but I still got questions, like how did that 36 in "36*36^x-1" vanish?
So with what you said here if I got for an example 5^x I can write it as 5*5^x-1 ?

this is like 5*5^x/5 ?

 

[Deleted member 4993]

Thanks this helps a lot, but I can't help but notice that you got 4^x-1 and 9^x-1 in the first case. Now I'm still wondering how that -1 ended up there?

When I take this post into consideration and try to solve this equation I get this:

Step-by-step:

9*16^x+4*81^x=13*36^x

9*(4^x)²+4*(9^x)²=13*(4*9)^x (or just 13*4^x*9^x)

4^x=u
9^x=v

9u²+4v²=13uv

9u²-13uv+4v²=0

(9u-4v)(u-v)=0

Case 1:
9u-4v=0
9*4^x=4*9^x



Case 2:
u-v=0
u=v
4^x=9^x




And from this point I'm not sure what to do since I am a bit rusty with logarithms, but I know that we can solve both these cases with logarithms and get the values you mentioned earlier.

Case 1:
9u-4v=0
9*4^x=4*9^x divide both sides by (9*4=)36 to get →

4^x/4 = 9^x/9

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

Since 2 and 3 are relatively prime, the above equation can be true only when:

x - 1 = 0

x = 1

Case 2:
u-v=0
u=v
4^x=9^x

Apply similar logic as above.

 

[ksdhart2]

Well this makes things a bit clearer but I still got questions, like how did that 36 in "36*36^x-1" vanish?

It didn't "vanish." We just divided both sides of the equation by 36. Taking another look at the equation in question, we have:

\({\color{red}9 \cdot 4} \cdot 4^{2x-1} + {\color{red}4 \cdot 9} \cdot 9^{2x-1} = {\color{red}36} \cdot 13 \cdot 36^{x-1}\)

\({\color{red}\cancel{9} \cdot \cancel{4}} \cdot 4^{2x-1} + {\color{red}\cancel{4} \cdot \cancel{9}} \cdot 9^{2x-1} = {\color{red}\cancel{36}} \cdot 13 \cdot 36^{x-1}\)

\(4^{2x-1} + 9^{2x-1} = 13 \cdot 36^{x-1}\)

 

[Steven G]

9*16^x + 4*81^x = 13*36^x

9*4*4^2x-1 + 4*9*9^2x-1 = 13*36*36^x-1

4^2x-1 + 9^2x-1 = 13*36^x-1

continue....

You really should see that x=0 is a solution as 9+4 does equal 13. Of course this does not help you see the other answers (if there are any)

To make things easier, I would use the following substitutions:

[MATH]u=4^x[/MATH]
[MATH]v=9^x[/MATH]
And then the equation becomes, when suitably arranged:

[MATH]9u^2-13uv+4v^2=0[/MATH]
Factor:

[MATH](9u-4v)(u-v)=0[/MATH]
Case 1:

[MATH]9u-4v=0[/MATH]
[MATH]4^{x-1}=9^{x-1}[/MATH]
Case 2:

[MATH]u-v=0[/MATH]
[MATH]4^{x}=9^{x}[/MATH]
In both cases, the equations are only true when the exponents are zero, meaning:

[MATH]x\in\{0,1\}[/MATH]

Outstanding work!

 

[Kalen12]

It didn't "vanish." We just divided both sides of the equation by 36. Taking another look at the equation in question, we have:

\({\color{red}9 \cdot 4} \cdot 4^{2x-1} + {\color{red}4 \cdot 9} \cdot 9^{2x-1} = {\color{red}36} \cdot 13 \cdot 36^{x-1}\)

\({\color{red}\cancel{9} \cdot \cancel{4}} \cdot 4^{2x-1} + {\color{red}\cancel{4} \cdot \cancel{9}} \cdot 9^{2x-1} = {\color{red}\cancel{36}} \cdot 13 \cdot 36^{x-1}\)

\(4^{2x-1} + 9^{2x-1} = 13 \cdot 36^{x-1}\)

Oh, okay, that's a bit clearer now, thanks.

 

[Kalen12]

You really should see that x=0 is a solution as 9+4 does equal 13. Of course this does not help you see the other answers (if there are any)

Outstanding work!

Well my biggest problem is writing down how I got to that solution, I know it's obvious that it's 0 but I have to prove it, if you know what I mean.

 

[Steven G]

Well my biggest problem is writing down how I got to that solution, I know it's obvious that it's 0 but I have to prove it, if you know what I mean.

No, I do not know what you mean. You simply notice that when you plug x=0 into the equation that the right hand side equals the left hand side.