If, log(x+y)/log2 = log(x–y)/log3 = log64/log0.125, find the values of x and y.
Answers
x = 22.63
y = 0.82
Explanation:
log(x+y)/log2 = log(x–y)/log3 = log64/log0.125
Taking each term separately, we get:
log(x+y)/log2 = log x + log y - log 2
log(x–y)/log3 = log x - log y - log 3
log64/log0.125 = log64 - log0.125
log(x–y)/log3 = log64/log0.125
log x - log y - log 3 - log 64 + log 0.125 = 0
log x + log y - log 2 - log 64 + log 0.125 = 0
log x - log y = log 3 + log 64 - log 0.125 ----------(1)
log x + log y = log 2 + log 64 - log 0.125----------(2)
2 log x = log 3 + log 2 + 2 log 64 - 2 log 0.125 -------------------- Adding (1) & (2)
-2 log y = log 3 - log 2 ----------------Subtracting (2) from (1)
log y² = log 2 - log 3 = log (2/3)
y² = 2/3
y = root of (2/3) = 0.82
2 log x = log 3 + log 2 + log 64 - log 0.125= log (3*2*64/0.125) = log 512
log x² = log 512
So x² = 512.
x = root of 512 = 22.63
Answer:
x=13/72
y=5/72
Explanation:
log 64/log 0.125
=>6 log 2/3 log (5/10)
=>2log 2/log (5/10)
=>2 log 2/log 5-log 5-log 2
=>2log 2/-log 2
=>-2
so log 64/ log 0.125 =-2
=>log (a+b)/log 2 =-2
=>log (a+b)=-2 log 2
=>log (a+b)= log (1/4)
=>a+b= 1/4 ..... (1)
,
=>log (a-b)/ log 3=-2
=>log (a-b) = -2 log 3
=>log (a-b) = log (1/9)
=>a-b= 1/9...(2)
from adding eq (1) and (2)
a+b= (1/4)
a-b= (1/9)
+---------------
=>2a=(1/4)+(1/9)
=>2a=13/36
=> a=13/72
from eq (1) we know that
a+b= 1/4
=>(13/72)+b=1/4
=>b=(1/4)-(13/72)
=>b=5/72
so a =13/72 and b =5/72 ANSWER...