If x=loga/log2a,y=log2a/log3a,z=log3a/log4a
Prove that xyz+1=2yz
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from given :
xyz + 1 = 2yz
loga/log2a × log2a/log3a × log3a/log4a + 1 = 2 × log2a/log3a × log3a/log4a
loga/ log4a + 1 = 2 × log2a/log4a
loga + log4a/log4a = 2 × log2a/log4a
loga + log4a = 2 × log2a
log(a × 4a) = 2 × log2a
log(2a)^2 = 2 × log2a
2 × log2a = 2 × log2a
L.H.S=R.H.S
xyz + 1 = 2yz
loga/log2a × log2a/log3a × log3a/log4a + 1 = 2 × log2a/log3a × log3a/log4a
loga/ log4a + 1 = 2 × log2a/log4a
loga + log4a/log4a = 2 × log2a/log4a
loga + log4a = 2 × log2a
log(a × 4a) = 2 × log2a
log(2a)^2 = 2 × log2a
2 × log2a = 2 × log2a
L.H.S=R.H.S
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