Question

Log(16/18)=x(log2-log3) then find the value of 2x-1

Answer 1

value of 2x - 1 would be 3 + log_(2/3) (2)²

it is given that,

log(16/18) = x(log2 - log3)

⇒log(8/9) = x( log2 - log3)

⇒log(2³/3²) = x(log2 - log3)

we know, log(m/n) = log m - log n

⇒log2³ - log3² = x (log2 - log3)

we also know, logaⁿ = nlog a

⇒3log2 - 2log3 = x(log2 - log3)

⇒x = (3log2 - 2log3)/(log2 - log3)

now 2x - 1 = 2{(3log2 - 2log3)/(log2 - log3) } - 1

= {6log2 - 4log3 - log2 + log3}/(log2 - log3)

= (5log2 - 3log3)/(log2 - log3)

= (log2^5 - log3³)/log(2/3)

= log(2^5/3³)/log(2/3)

= log(2³ × 2²)/3³)/log(2/3)

= {log(2/3)³ + log(2)²}/log(2/3)

= {3log(2/3) + log(2)²}/log(2/3)

= 3 + log_(2/3)2²

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Answer 2

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it is given that,

log(16/18) = x(log2 - log3)

⇒log(8/9) = x( log2 - log3)

⇒log(2³/3²) = x(log2 - log3)

we know, log(m/n) = log m - log n

⇒log2³ - log3² = x (log2 - log3)

we also know, logaⁿ = nlog a

⇒3log2 - 2log3 = x(log2 - log3)

⇒x = (3log2 - 2log3)/(log2 - log3)

now 2x - 1 = 2{(3log2 - 2log3)/(log2 - log3) } - 1

= {6log2 - 4log3 - log2 + log3}/(log2 - log3)

= (5log2 - 3log3)/(log2 - log3)

= (log2^5 - log3³)/log(2/3)

= log(2^5/3³)/log(2/3)

= log(2³ × 2²)/3³)/log(2/3)

= {log(2/3)³ + log(2)²}/log(2/3)

= {3log(2/3) + log(2)²}/log(2/3)

= 3 + log_(2/3)2²