If log 1218 - - and log 2454 - - - prove that - -5- - - -1

Log x=log(2× 3^2)=log 2+3log 3 log y=2log 2+log 3 a=log xlog y b=log 3+log xlog 2+log y ab+5(ab)=log× (log 3+log x)log y(log 2+log y)+5(log xlog y) log ...

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Byju's Answer

Standard XII

Mathematics

Algebra of Complex Numbers

If log 1218...

Question

If $l o g_{12} 18 = α$ and $l o g_{24} 54 = β$ , prove that $α β + 5 (α - β) = 1$

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Solution

$l o g x = l o g (2 \times 3^{2}) = l o g 2 + 3 l o g 3$
$l o g y = 2 l o g 2 + l o g 3$
$a = \frac{l o g x}{l o g y}$
$b = \frac{l o g 3 + l o g x}{l o g 2 + l o g y}$
$a b + 5 (a b) = \frac{l o g \times (l o g 3 + l o g x)}{l o g y (l o g 2 + l o g y)} + 5 (\frac{l o g x}{l o g y}) - \frac{l o g 3 + l o g x}{l o g 2 + l o g y}$
$= \frac{5 l o g 2 l o g 3 + (l o g 3)^{2} + 6 (l o g 2)^{2}}{(2 l o g 2 l o g 3) (3 l o g 2 + l o g 3)}$
$= \frac{5 l o g 2 l o g 3 + (l o g 3)^{2} + 6 (l o g 2)}{5 l o g 2 l o g 3 + (l o g 3)^{2} + 6 (l o g)^{2}}$
$= 1$ .

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If log1218=α and log2454=β , prove that αβ+5(α−β)=1

logx=log(2×32)=log2+3log3logy=2log2+log3a=logxlogyb=log3+logxlog2+logyab+5(ab)=log×(log3+logx)logy(log2+logy)+5(logxlogy)−log3+logxlog2+logy=5log2log3+(log3)2+6(log2)2(2log2log3)(3log2+log3)=5log2log3+(log3)2+6(log2)5log2log3+(log3)2+6(log)2=1.

If $l o g_{12} 18 = α$ and $l o g_{24} 54 = β$ , prove that $α β + 5 (α - β) = 1$