Posted by Katie on Tuesday, March 5, 2013 at 10:59pm.
assuming logs base 10,
2logx = log_6(8)
logx = 1/2 log_6(8)
x = √(10^(log_6(8))
now log_6(8) = log6/log8. so
x = √(10^log6)^(1/log8)
= √(6^1/log8)
= 6^(1/log64)
take log of both sides
log [ 6^(2logx) ] = log8
2logx (log6) = log8
2logx = log8/log6
log x^2= log8/log6
x^2= 10^(log8/log6)
x = √( 10^(log8/log6) ) = 3.804339012
check using your calculator, it works
Go with Reiny's answer. I messed up my log base conversion.
Related Questions
algebra - equivalent expression for 2log(x+1)-2log(x-1)-log(x-1)
Math - Express as a single logarithm. 1/2log(subscript)3 144-log(subscript)3 4+...
algebra - does log base of 4 sqrt(t/s)=1/2log base of 4 t - 1/2log base of 4s? ...
Precalculus - What am I doing wrong? I keep getting 48 and its wrong... Solve ...
math - how do you Solve 2log x-log 3=2 pls help
Math - Are you allowed to do this 2log(x-1) =log(x+1) 2 (x-1)=(x+1) x=3 Please ...
LOGARITHMS - HI THE QUESTION STATES THAT ONE SHOULD CALCULATE THE LOGS WITHOUT ...
math - also a problem that i cant figure out is how to do inverses of logs f(x...
math - solve for x 32^x+5=16^2x+1 1/2log(x+2)=2 sqrt(2x-3)-2=5
Math: Logarithms - (log(x^6)+2log(y^3))/log(xy)
For Further Reading
