SOLUTION: Solve for x in the following eguation. 5=9^x-1 (as in x-1 are exponents.) I can't figure out where the log goes... :(

Algebra ->  Algebra  -> Exponential-and-logarithmic-functions -> SOLUTION: Solve for x in the following eguation. 5=9^x-1 (as in x-1 are exponents.) I can't figure out where the log goes... :(      Log On

Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help!
Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations!

   

Question 71846: Solve for x in the following eguation.
5=9^x-1
(as in x-1 are exponents.)
I can't figure out where the log goes... :(
Answer by jim_thompson5910(21685) About Me  (Show Source):
You can put this solution on YOUR website!
The log of base 9 goes on both sides. The log undoes an exponent, so we use the log on the entire right side, but we must do it to both sides.
5=9%5E%28x-1%29
log_%5B9%5D%285%29=log_%5B9%5D%289%5E%28x-1%29%29Take the log of both sides.The log simply cancels out the base 9
log_%5B9%5D%285%29=+x-1Add 1 to both sides
log_%5B9%5D%285%29%2B1=xSince log_[9](5) is a number we can find x
Use the change of base formula
log_%5Bb%5D%28x%29=log%28x%29%2Flog%28b%29Where b is the base and x is the argument. Remember the base of the log is assumed to be 10 if not specified.
x=log%285%29%2Flog%289%29%2B1So this can be evaluated on a calculator
x=0.69897%2F0.95424+%2B+1
x=0.73248+%2B+1
x=1.73248
Check:
5=9%5E%281.73248-1%29
5=9%5E%280.73248%29
5=4.99993 Which is close enough to 5 (there are some roundoff errors).