# SOLUTION: I do apolgize. I just submitted a problem but I guess I didn't write it correctly. I'm still very new at this. the question is 4^(x+1)=64 the x+1 are in exponential fo

Algebra ->  Algebra  -> Logarithm Solvers, Trainers and Word Problems -> SOLUTION: I do apolgize. I just submitted a problem but I guess I didn't write it correctly. I'm still very new at this. the question is 4^(x+1)=64 the x+1 are in exponential fo      Log On

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 Click here to see ALL problems on logarithm Question 22799: I do apolgize. I just submitted a problem but I guess I didn't write it correctly. I'm still very new at this. the question is 4^(x+1)=64 the x+1 are in exponential form. Where appropriate, include approximations to the nearest thousandth. Again thank you so much. Found 2 solutions by stanbon, khwang:Answer by stanbon(57307)   (Show Source): You can put this solution on YOUR website!I'm going to assume you do not want to solve this problem using logarithms. If you do the procedure is quite different. Without using logs you proceed as follows. Rewrite the equation as follows: 4^(x+1) = 4^3 Then x+1 = 3 x = 4 Hope this helps. Cheers, Stan H. Answer by khwang(438)   (Show Source): You can put this solution on YOUR website! x=2 instead and x^-y^2 = 53 is a hyperboa. Kenny