SOLUTION: I am trying to solve the following equation:
log_5 x/125 = log_7(49x^3)
This is what I have tried...where am I going wrong?
log_5 (x) - log_5 (5^3) = log_7(7^2) + log_7 (x
Algebra.Com
Question 75928: I am trying to solve the following equation:
log_5 x/125 = log_7(49x^3)
This is what I have tried...where am I going wrong?
log_5 (x) - log_5 (5^3) = log_7(7^2) + log_7 (x^3)
log_5 (x) - 3 = 2 + 3 log_7 (x)
-5 = x(3 log_7 - log_5)
Please help....when I now divide...I get x to be undefined?
Thank you
Answer by scott8148(6628) (Show Source): You can put this solution on YOUR website!
You're fine until the last step...you "factor out" the argument of the two logs, leaving undefined values (log_7 and log_5 of what?)...You need to get the bases of the logs the same (like common denominators for fractions).
-5=3log_7(x)-log_5(x)... this equals... all bases now the same
multiplying by log 7 and log 5 and factoring gives...-5 log7 log5 = logx (3 log5 - log7)...dividing by (3 log5 - log7) gives ...so x=0.00437...
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