Expert: Sherry Wallin - 5/24/2010

Question
Log9x + 3log3x = 14.  9 and 3 are bases

Answer
Sorry it has taken me so long to repsond but I didn't have internet access, just my celly.Anyway here is the answer I prepared:

When doing  a log problem with different bases you need to change each into the same base. There is a change of base formula that says: log_b(x) = log_d(x)/log_d(b) where the _ indicates the base. What this is saying (in words) is that you can rewrite a log in any base as the log of the argument divided by the log of the base. The simplest thing to do of course is to let d = 10 or d = e  so that you can calculate these on your calculator. So take log_9 x and rewrite it as log_10 (x)/log_10  (9) where I put parenthesis around the functions argument and showed I was changing the base 9 to a base 10 problem by using _10 for base 10. Now change the other term to base 10:
3Log_3 (x) = 3[log_10 (x)/log_10 (3)]
So the complete problem is now:  log_9(x) + 3log_3(x) = 14 ->
log_10 (x)/log_10  (9) + 3[log_10 (x)/log_10 (3)] = 14
now since log_10(9) and log_10(3) is just a number you could replace them with their real number representation:  log_10 (9) ~ .954 and log_10(3) ~.477 so…
log_10 (x)/log_10  (9) + 3[log_10 (x)/log_10 (3)] = 14 -> log_10(x)/.954 +3log_10(x)/.477 = 14 ->
log_10(x)[1/.954 + 3*(1/.477)] = 14 -> log_10(x) [1.048 + 6.289] = 14 ->
log_10(x) = 14/(1.048 + 6.289)-> log_10(x) = 14/7.337 -> log_10(x) = 1.908 -> 10^(1.908) ~ 80.91

Math Prof