SOLUTION: Here is my equation: 4^x + 6 = 7 Here is what I tried: x + 6log4 = 7log x (log4 + 6log4) = 7log x = 7log/log4 + 6log4 I come up with 3.61 but I dont think that is correct.

Algebra ->  Algebra  -> Exponential-and-logarithmic-functions -> SOLUTION: Here is my equation: 4^x + 6 = 7 Here is what I tried: x + 6log4 = 7log x (log4 + 6log4) = 7log x = 7log/log4 + 6log4 I come up with 3.61 but I dont think that is correct.       Log On

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Question 22516: Here is my equation: 4^x + 6 = 7
Here is what I tried: x + 6log4 = 7log
x (log4 + 6log4) = 7log
x = 7log/log4 + 6log4
I come up with 3.61 but I dont think that is correct.
Here is the other way I was trying to solve it.
ln4^x + 6 = ln 7
x + 6 ln 4 = ln 7
x = ln7/6ln4 = .45 Is that way correct?
Answer by ilana(305) About Me  (Show Source):
You can put this solution on YOUR website!
Actually, you are making this much more difficult than necessary. 4^x is just a term, so pretend it's 4x for now. Then we have 4x+6=7. You then subtarct 6 from both sides to get 4x=1, so x=1/4. In your problem, begin the same way but get 4^x=1. (Any number)^0=1, so x=0.