SOLUTION: How do you get the solution to {{{4^(x+1)=12}}} Do you put x+1 In the front of 4 Then use ln?

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: How do you get the solution to {{{4^(x+1)=12}}} Do you put x+1 In the front of 4 Then use ln?      Log On


   

Question 673981: How do you get the solution to 4%5E%28x%2B1%29=12
Do you put x+1 In the front of 4
Then use ln?
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
4%5E%28x%2B1%29=12..........since 4%5E%28x%2B1%29=4%5Ex%2A4 replace 4%5E%28x%2B1%29 with 4%5Ex%2A4
4%5Ex%2A4=12
4%5Ex=3...now use a logarithm
log%28a%2C%28x%29%29=+N... means that in your case a%5EN+=+x.....a=4, N=x ,and x=3; so, we have
log%284%2C%28+3+%29%29=+x.......transfer into log base 10

log%283%29%2Flog%284%29=+x
since log%283+%29+=+1.0986122886681098 and log%284%29=+1.3862943611198906188344642429163531361510002687205105, we have

x=0.7924812503605781690686400288008583335496142251280518...round it to two decimals
x=0.79