Help with Logarithms: solving 4^(x^2 + 2) = 64^x

[mrtentaclenun]

Hello! My first time posting. I am having the hardest time solving this math problem. I have tried different approaches and asked others for help but no one seems to be able to give me the right kind of help. The directions say "Solve each equation for x. (Express the answer in exact form, then give the decimal approximation to four decimal places.)"
The problem I am having issues with is

4^(x^2 + 2) = 64^x

It seems simple and yet I just cant figure it out. Thank you!

 

[wjm11]

Re: Help with Logarithms

4^(x^2 + 2) = 64^x

The key here is to get the base numbers to match on both sides of the equation. Rewrite 64 as 4^3. Can you take it from here?

 

[fasteddie65]

Just in case you don't:

4^(x^2 + 2) = 64^x
4^(x^2 + 2) = (4^3)^x = 4^(3x)
x^2 + 2 = 3x
x^2 - 3x + 2 = 0
(x - 1)(x - 2) = 0
x = 1 or x = 2

 

[mmm4444bot]

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