Exponential Troubles!

[jkovie]

HI, I have one problem that I can't seem to get to the answer and I know the answer, but I don't know how to get it.

This is the Problem: Solve for X: 4^(5x) - 2^(3x-5) = 8^9

I made similar bases:

2^(10x) - 2^(3x-5) = 8^9

Our answer key says the answer is X = 11 but I have no clue how they got it. Thanks for helping.

 

[soroban]

Hello, jkovie!

Is there a typo?
Their answer is incorrect.

\(\displaystyle \text{Solve for }x\!:\;\;4^{5x} - 2^{3x-5} \:=\: 8^9\)

\(\displaystyle \text{Answer key: }\: x = 11\) . No

If \(\displaystyle x = 11\), we have:.\(\displaystyle 4^{55} - 2^{28}\) . . . a 34-digit number

. . which is not equal to \(\displaystyle 8^9 \,=\,134,217,728.\)

 

[jkovie]

Thank you Soroban, Exactly. 11 does not work, so that has to be incorrect. I still am not sure how to proceed to figure for X however. If you could steer me in the right direction, I would appreciate it.

Solve for X: 2^(10x) - 2^(3x-5) = 8^9

Do I need to use Logs or can I somehow use the Exponents to solve?

Thanks.

 

[jkovie]

I just figured it out but it still doesn't make sense for X = 11

Solve for X : 4^5x - 2^(3x-5) = 8^9

Can you take the exponents on the left: 5x - (3x-5) and end up with 2x+5 on the left

Then break 8^9 to (2^3)^9 = 2^27

then have 2x+5 = 27 Solving for x = 11

I get the 11 but to plug it into the equation still does not fit the number for 8^9 either.

This one is confusing.

 

[srmichael]

I just figured it out but it still doesn't make sense for X = 11

Solve for X : 4^5x - 2^(3x-5) = 8^9

Can you take the exponents on the left: 5x - (3x-5) and end up with 2x+5 on the left

Then break 8^9 to (2^3)^9 = 2^27

then have 2x+5 = 27 Solving for x = 11

I get the 11 but to plug it into the equation still does not fit the number for 8^9 either.

This one is confusing.

No, you can't do that. Wolfram gets x approximately equal to 2.7

 

[jkovie]

Thanks Smichael. I am going to recheck this equation, because the answer key does not work in the equation.

But as far as my steps, am I doing it correctly?

 

[Deleted member 4993]

That answer would make sense if your equation were:

\(\displaystyle \dfrac{2^{5x}}{2^{3x-5}} \ = \ 8^9 \)

or

2^5x ÷ 2^3x-5 = 8⁹

 

[srmichael]

Thanks Smichael. I am going to recheck this equation, because the answer key does not work in the equation.

But as far as my steps, am I doing it correctly?

Yes, your first step (\(\displaystyle 4^{5x}-2^{3x-5}=8^9\)) is correct, but you can actually go a little further:

\(\displaystyle 4^{5x}-2^{3x-5}=8^9\)

\(\displaystyle 2^{2^{5x}}-2^{3x-5}=2^{3^9}\)

\(\displaystyle 2^{10x}-2^{3x-5}=2^{27}\)

 

[HallsofIvy]

Thanks Smichael. I am going to recheck this equation, because the answer key does not work in the equation.

But as far as my steps, am I doing it correctly?

No, you are not. You can't just "take the exponents" because you have different bases: \(\displaystyle 3^x= 5^y\) does NOT mean "x= y"!

However you can use the fact that \(\displaystyle 4= 2^2\) to replace \(\displaystyle 4^{5x}\) with \(\displaystyle (2^2)^{5x}= 2^{10x}\) and, because \(\displaystyle 8= 2^3\), \(\displaystyle 8^9= (2^3)^9= 2^{27}\). That is \(\displaystyle 4^{5x}- 2^{3x- 5}= 8^9\) becomes \(\displaystyle 2^{10x}- 2^{3x- 5}= 2^{27}\).

You still cannot just "take the exponents". You need to find some way to write it as \(\displaystyle 2^a= 2^b\). Then you can say "a= b".

 

[lookagain]

Yes, your first step (\(\displaystyle 4^{5x}-2^{3x-5}=8^9\)) is correct, but you can actually go a little further:

\(\displaystyle 4^{5x}-2^{3x-5}=8^9\)

\(\displaystyle 2^{2^{5x}}-2^{3x-5}=2^{3^9}\) < < < This is incorrect. The second line does not follow from the first line. The exponents are not handled this way. Grouping symbols are required.


\(\displaystyle 2^{10x}-2^{3x-5}=2^{27}\)

The second line should be:


\(\displaystyle (2^2)^{5x} - 2^{3x-5} = (2^3)^9\)



In general, for appropriate a, b, and c, we have that


\(\displaystyle a^{b^{c}} = a^{(b^c)}, \ not \ \ the \ \ desired \ \ (a^b)^{c}.\)