SOLUTION: 4^2x+1=64 4^log X=8 e^x+1=.05 Please help me solve these problems. Thank you!

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: 4^2x+1=64 4^log X=8 e^x+1=.05 Please help me solve these problems. Thank you!      Log On


   

Question 186471: 4^2x+1=64
4^log X=8

e^x+1=.05
Please help me solve these problems. Thank you!
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
4^(2x+1) = 64
4^(2x+1) = 4^3
2x+1 = 3
2x = 2
x = 1
------------------------
4^log X=8
2^(2logx) = 2^3
2logx = 3
logx = 3/2
x = 10^(3/2)
x = 31.6228
-------------------------
e^(x+1)=.05
Take the natural log of both sides to get:
x+1 = ln(0.5)
x+1 = -2.9957
x = -3.9957
----------------------------
Cheers,
Stan H.