Given the simultaneous equations

2^x=3^y

x+y=1

show that x=log3/log6

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- May 21st 2006, 02:05 AMkingkaisai2Difficult A2 logarithm
Given the simultaneous equations

2^x=3^y

x+y=1

show that x=log3/log6 - May 21st 2006, 03:46 AMticbolQuote:

Originally Posted by**kingkaisai2**

x +y = 1 ---------(2)

Take the logs of both sides of (1),

x*log(2) = y*log(3) ----(1a)

From (2),

y = 1-x

Substitute that into (1a),

x*log(2) = (1-x)log(3)

x*log(2) = log(3) -x*log(3)

x*log(2) +x*log(3) = log(3)

x[log(2) +log(3)] = log(3)

x[log(2*3)] = log(3)

x*log(6) = log(3)

x = log(3) / log(6) ---------answer. - May 30th 2006, 05:41 AMmalaygoelQuote:

Originally Posted by**kingkaisai2**

taking log on both sides of the first equation

xlog2=ylog3

substituting y from second equation

xlog2=(1-x)log3

rearranging

x(log2+log3)=log3

therefore,x=log3/log6