Expert: Sherry Wallin - 12/29/2009

Question
1.) 9^(3x+1)= 7^x

2.) log(x-2)-log(x+4)=5

3.) log6(x+1)=log3(x-8) 6 and 3 are the base of their logs*

I am unsure how to solve for x in either of these three scenarios and was hoping you could help me figure out how to do them.

Answer
When the bases are not the same and they cannot be made the same then you have no choice but to take the log of(ln) of both sides and then solve like this:

1-

9^(3x+1)= 7^x
log[9^(3x+1)]=log(7^x)
(3x+1)log(9) = xlog(7) using the rule rlog x = log r^x
3x*log(9) + log(9) = x*log(7) distributing
3xlog(9)-xlog(7) = -log(9)  gather terms with a factor of x on one side
x[3log(9) - log(7)] = -log(9)  factor x out
x = -log(7)/[3log(9)-log(7)] get x by itself
x = -log(7)/[log(9^3)-log(7)] rule for rlog x = log r^x
x = -log(7)/[log(729)/log(7)]  the difference of the logs is the log of a quotient
x = -log(7)*log(7)/log(729) simplification of a fraction through division
x = log(7^-1)*log(7)/log(729)  rule rlog x = log r^x
x = log (1/7)*log(7)/log(729)

2-

log(x-2)-log(x+4)=5
log[(x-2)/(x+4)]= 5   log of the difference is the log of the quotient
10^5 = (x-2)/(x+4) the base of log is 10, and using the rule log_(10) x = y if and only if 10^y = x
10^5(x+4) = x - 2 cross multiply
10^5*x +4*10^5 = x - 2
100,000x - x = - 400,000-2
99,999x = -400,002
x = -400,002/99,999 which is approximately -4 which means there is no solution since you can only
take the log of a positive number and both log(x-2)-log(x+4) are undefined for x = -4

3-  

log6(x+1)=log3(x-8)  use the rule Log_a(M) = log M/log a
[log(x+1)/log 6]= [log(x-8)/log 3] now they are both base 10
log(x+1)*log(3) = log(x-8)*log(6) cross multiply
                                 log(3) and log(6) are just numbers call then u and v respectively
.477log(x+1) ~= .778log(x-8)
log(x+1)^.477 ~= log(x-8)^.778
log(x+1)^.477 - log(x-8)^.778 = 0
log[(x+1)^.477/log(x-8)^.778] = 0
10^0 = [(x+1)^.477/(x-8)^.778]
1 = [(x+1)^.477/(x-8)^.778]
and this is tedious, so if you want to solve for x use algebra and do it...

Math Prof