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Logarithms. Calculate the value of y for which 2log 3 y -log 3 ( y +4)=2. 2log 3 y -log 3 ( y +4)=2 log 3 y 2 -log 3 ( y +4)=2 log 3 ( y 2 /y +4)=2 y 2 /y +4=3 2 y 2 /y +4=9 y 2 =9( y +4) y 2 -9 y -36=0 ( y -12)( y +3)=0. Use the technique that log(x)-log(y)=log(x/y).
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Calculate the value of y for which 2log3y-log3(y+4)=2 2log3y-log3(y+4)=2 log3y2-log3(y+4)=2 log3(y2/y+4)=2 y2/y+4=32 y2/y+4=9 y2=9(y+4) y2-9y-36=0 (y-12)(y+3)=0 Use the technique that log(x)-log(y)=log(x/y)
Calculate the value of z for which log3z=4logz3 log3z=4logz3 1/logz3=4logz3 1=4(logz3)2 ¼=(logz3)2 ½=±logz3 z½=3 or z=3 z=9 Use the technique that logy(x)=1/logx(y)
Solve the simultaneous equations8y=42x+3 and log2y=log2x+4 8y=42x+3 (23)y=(22)2x+3 23y=24x+6 3y=4x+6 log2y=log2x+4 log2y=log2x+log216 log2y=log2(x×16) y=16x 3y=48x 3y=4x+6 0=44x-6 x=3/22 Use the technique that log(x)+log(y)=log(xy)
Solve the equation log3(2-3x)=log9(6x2-19x+2) Use the technique that logy(x)=logq(x)/ logq(y) log3(2-3x)=log9(6x2-19x+2) log3(2-3x)=[log3(6x2-19x+2)]/log3(9) log3(2-3x)=[log3(6x2-19x+2)]/2 2log3(2-3x)=log3(6x2-19x+2) log3(2-3x)2=log3(6x2-19x+2) (2-3x)2=6x2-19x+2 4-12x+9x2=6x2-19x+2 2+5x+3x2=0 2+2x+3x+3x2=0 2(1+x)+3x(1+x)=0 (2+3x)(1+x)=0 x=-2/3 x=-1