SOLUTION: Hello I need help with this exponential equation
2^2x times 4^4x+8 = 64
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Question 804377: Hello I need help with this exponential equation
2^2x times 4^4x+8 = 64
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
I presume the "+ 8" is in the exponent. Please use parentheses in the future so that I don't have to guess what you mean. Parentheses would have obviated the need to spell out "times" also.
\left(4^{4x\,+\,8}\right)\ =\ 64)
Note that ^m\ =\ b^{nm})
, so 
And 
, so:
(4^{4x+8})\ =\ 4^{5x\,+\,8}\ =\ 64)
Take the base 4 log of both sides:
\ =\ \log_4(64))
But \ =\ n\log_b(x))
, and since 
we get:
\log_4(4)\ =\ 3)
Then since \ =\ 1)
(See why I picked base 4 for the log?)
John
Egw to Beta kai to Sigma
My calculator said it, I believe it, that settles it
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