SOLUTION: 2log[4](3x)+log[4](11)=14
Solve. I don't know how to. The answer I could come up with is
2log[4](3)+2log[4](x)+log[4](11)=14
But I don't think that's right. Is it?
Algebra ->
Exponential-and-logarithmic-functions
-> SOLUTION: 2log[4](3x)+log[4](11)=14
Solve. I don't know how to. The answer I could come up with is
2log[4](3)+2log[4](x)+log[4](11)=14
But I don't think that's right. Is it?
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Question 618258: 2log[4](3x)+log[4](11)=14
Solve. I don't know how to. The answer I could come up with is
2log[4](3)+2log[4](x)+log[4](11)=14
But I don't think that's right. Is it? Answer by lwsshak3(11628) (Show Source):
You can put this solution on YOUR website! 2log[4](3x)+log[4](11)=14
place under single log
log4[(3x)^2*11]=14
convert to exponential form: base(4) raised to log of number(14)=number[(3x)^2*11]
4^14=[(3x)^2*11]=99x^2
x^2=4^14/99
x=√(4^14/99)≈2711469.25
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