Asked by Matteo
The cariable x satisfies the equation 3^x.4^(2x+1)=6^(x+2).
By taking logarithms of both sides,show that x=log9/log8
By taking logarithms of both sides,show that x=log9/log8
Answers
Answered by
Reiny
log 3^x + log 4^(2x+1) = log 6^(x+2)
xlog 3 + (2x+1)log 4 = (x+2)log 6
xlog 3 + 2xlog 4 + log 4 = xlog 6 + 2log 6
x(log3 + 2log4 - log6) = 2log6 - log4
x(log3 + log16 - log6) = log36 - log4
x(log(3x16/6) = log(36/4)
x(log 8) = log 9
x = log9/log8 as required.
xlog 3 + (2x+1)log 4 = (x+2)log 6
xlog 3 + 2xlog 4 + log 4 = xlog 6 + 2log 6
x(log3 + 2log4 - log6) = 2log6 - log4
x(log3 + log16 - log6) = log36 - log4
x(log(3x16/6) = log(36/4)
x(log 8) = log 9
x = log9/log8 as required.
Answered by
Matteo
wow thanks soooo much!!!!! Great help :)
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