SOLUTION: Hi, could you please help me to solve the following equation? log(3,x).log(9,x).log(27,x)=4/3

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Question 202713: Hi, could you please help me to solve the following equation?
log(3,x).log(9,x).log(27,x)=4/3
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
If the problem is not
log%283%2C+x%29%2Alog%289%2C+x%29%2Alog%2827%2C+x%29+=+4%2F3
then what are the "." separators?

To solve the problem we need to rewrite each log in the same base. Since 27 and 9 are powers of 3 it will not be hard to rewrite each log with a base of 3.

Let y+=+log%2827%2C+x%29
Rewriting this in exponential form:
27%5Ey+=+x
Substituting 3%5E3 for 27:
%283%5E3%29%5Ey+=+x
Using the %28a%5Eb%29%5Ec+=+a%5E%28b%2Ac%29 rule for exponents we get:
3%5E%283y%29+=+x
Finding the log base 3 of both sides:
log%283%2C+%283%5E%283y%29%29%29+=+log%283%2C+x%29
Using the log%28%28a%5Eb%29%29+=+b%2Alog%28%28a%29%29 property of logarithms we get:
3y%2Alog%283%2C+3%29+=+log%283%2C+x%29
Since log%283%2C+3%29+=+1 we get
3y+=+log%283%2C+x%29
Multiplying both sides by 1/3 we get:
y+=+%281%2F3%29log%283%2Cx%29
Substituting back in for y we get:
log%2827%2C+x%29+=+%281%2F3%29log%283%2C+x%29
Using similar logic on log%289%2C+x%29 we get:
log%289%2C+x%29+=+%281%2F2%29log%283%2C+x%29
Substituting these two converted logs into the original equation we get:
log%283%2C+x%29%2A%281%2F2%29log%283%2C+x%29%2A%281%2F3%29log%283%2C+x%29+=+4%2F3
which simplifies to:
%281%2F6%29%28log%283%2C+x%29%29%5E3+=+4%2F3
Multiplying both sides by 6 we get:
%28log%283%2C+x%29%29%5E3+=+8
Finding the cude root of both sides we get:
log%283%2C+x%29+=+2
Rewriting this in exponential form:
x+=+3%5E2+=+9
which is our solution.