SOLUTION: what is the way to solve the logarithms a.) 2e^9x=4 b.)ln(4x − 3) = 4 and c.) 7^x + 1 = 4? I keep getting a negative answer fot c. and I know it is not supposed to be that

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: what is the way to solve the logarithms a.) 2e^9x=4 b.)ln(4x − 3) = 4 and c.) 7^x + 1 = 4? I keep getting a negative answer fot c. and I know it is not supposed to be that      Log On


   

Question 599931: what is the way to solve the logarithms a.) 2e^9x=4 b.)ln(4x − 3) = 4 and c.)
7^x + 1 = 4? I keep getting a negative answer fot c. and I know it is not supposed to be that way. I am getting depressed. Can you please help? ex.:
7^x+1=4, log7^x+1=4, (x+1)log7=log4, xlog7+log7=log4-log7, x=log4-log7/log7. All my answers are negatives, and are wrong.
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
a.) 2e^9x=4
e^9x=4/2=2
take log of both sides
9xlne=ln2
lne=1
9x=ln2
x=ln2/9
x≈0.077
..
b.)ln(4x − 3) = 4
convert to exponential form: base(e) raised to log of number(4)=number(4x-3)
e^4=4x-3
4x=e^4+3
x=(e^4+3)/4
x≈14.4
..
c.)7^x + 1 = 4
7^x=3
xlog7=log3
x=log3/log7
x≈0.564