Question

if a=log12 18 and b=log24 54 then find the value of ab+5(a-b)

Answer 1

Given,
$\bold{a=log_{12}^{18}}\\\\\bold{b=log_{24}^{54}}$
we have to find out ab + 5 (a - b)
= $\bold{log_{12}^{18}.log_{24}^{54}+5(log_{12}^{18}-log_{24}^{54})}$

log₁₂¹⁸ = log18/log12 = log(3² × 2)/log(2² × 3)
= {log3² + log2}/{log2² + log3}
= (2log3 + log2)/(2log2 + log3)
similarly we have to resolve log₂₄⁵⁴ = (3log3 + log2)/(3log2 + log2)
Let log3 = x and log2 = y
Then, log₁₂¹⁸ = (2x + y)/(2y + x )
log₂₄⁵⁴ = (3x + y)/(3y + x)

Now, $\bold{a=log_{12}^{18}.log_{24}^{54}+5(log_{12}^{18}-log_{24}^{54})}$
= (2x + y)/(2y + x).(3x + y)/(3y+x)+ 5[(2x + y)/(2y + x) - (3x + y)/(3y + x) ]
= (2x + y)/(2y + x).(3x + y)/(3y + x) + 5[(2x + y)(3y + x) - (3x + y)(2y + x)]/(2y + x)(3y + x)
= (6x² + 5xy + y²)/(2y + x)(3y + x) + 5[(7xy + 2x² + 3y² - 7xy - 3x² - 2y² ]/(2y + x)(3y + x)
= (x² + 5xy + 6y²)/(2y + x)(3y + x)
= (x + 3y)(x + 2y)/(2y + x )(3y + x)
= 1

Hence, answer is 1

Answer 2

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