plzzz ans.....Thnk you ......Logarithms
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Hey there !!!
logxy=log(x/y)+2log2=2
logxy=logx+logy log(x/y)=logx-logy
So,
logx+logy=logx-logy+2log2=2
logx+logy=2-------Equation 1
logx-logy+2log2=2
logx-logy=2-2log2
logx-logy=2(1-log2)-------Equation 2
Adding equation 1 and 2
logx+logy=2
+(logx-logy=2(1-log2))
___________________
2logx=2+2(1-log2)
2logx=2(1+1-log2)
logx=2-log2
We know that log10=1
similarly log100=log10²=2log10=2
So, 2=log100
Now,
logx=2-log2
logx=log100-log2
logx=log(100/2)
logx=log50
⇒⇒x=50
logxy=2
logxy=log100
xy=100
But x=50
So, 50*y=100
y=100/50=2
So, x=50 y=2
Hope this helped you.................
logxy=log(x/y)+2log2=2
logxy=logx+logy log(x/y)=logx-logy
So,
logx+logy=logx-logy+2log2=2
logx+logy=2-------Equation 1
logx-logy+2log2=2
logx-logy=2-2log2
logx-logy=2(1-log2)-------Equation 2
Adding equation 1 and 2
logx+logy=2
+(logx-logy=2(1-log2))
___________________
2logx=2+2(1-log2)
2logx=2(1+1-log2)
logx=2-log2
We know that log10=1
similarly log100=log10²=2log10=2
So, 2=log100
Now,
logx=2-log2
logx=log100-log2
logx=log(100/2)
logx=log50
⇒⇒x=50
logxy=2
logxy=log100
xy=100
But x=50
So, 50*y=100
y=100/50=2
So, x=50 y=2
Hope this helped you.................
sanj2:
thnk i
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