SOLUTION: I'm just wondering if it's possible to simplify this further or if I'm even on the right track... 3^x=15 (find x) >x=log₃15 >x=log₃(3x5) >x=log₃3 + log&#8

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: I'm just wondering if it's possible to simplify this further or if I'm even on the right track... 3^x=15 (find x) >x=log₃15 >x=log₃(3x5) >x=log₃3 + log&#8      Log On


   

Question 255028: I'm just wondering if it's possible to simplify this further or if I'm even on the right track...
3^x=15 (find x)
>x=log₃15
>x=log₃(3x5)
>x=log₃3 + log₃5
>x= 1 + log₃5
Thanks for your help :o)
Found 2 solutions by Greenfinch, Alan3354:
Answer by Greenfinch(383) About Me  (Show Source):
You can put this solution on YOUR website!
Second line should be
x log3 = log 15 giving
x = log 15 / log 3 = 1.1770 / 0.4771
I haven't got a calculator handy but it is 2. something which fits since x = 2 would give 3^2 = 9 and 3^3 = 27

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
3^x=15 (find x)
>x=log₃15
>x=log₃(3x5)
>x=log₃3 + log₃5
>x= 1 + log₃5
= 1 + (log(5)/log(3))
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Second line should be
x log3 = log 15 giving
x = log 15 / log 3 = 1.1770 / 0.4771
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The result is the same, but Greenfinch's method is faster and simpler.
= 2.46497...