Kalambo Junior
Senior Member
- Jul 15, 2011
- 127
- 23
yes! it's correct2^x=4x apply log each side log2^x=log4x xlog2=log4+logx xlog2=2log2+logx xlog2-2log2=logx log2(x-2)=logx dividing by log2(x-2) throughout the eqn we obtain 1=logx/log2(x-2) this can also be writen as logxbase2(x-2)=1, into exponential form it will be 2(x-2)^1=x 2(x-2)=x 2x-4=x x=4 ok!