JEE Novice - (23)

Algebra Level 3

$\large \begin{cases} \log_{30} 3 & = & a \\ \log_{30} 5 & = & b \\ \log_{30} 16 & = & ? \end{cases}$

Give your answer in terms of $a$ and $b$ .

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4(1+a+b)

4(1-a-b)

4ab

4(1-a+b)

3 solutions

Manish Dash
May 27, 2015

$We\quad know,\quad \log _{ 30 }{ 3 } =\quad a,\quad \quad \log _{ 30 }{ 5 } =\quad b\\ \\ \therefore \quad \quad \log _{ 30 }{ 3 } +\log _{ 30 }{ 5 } \quad =\quad \log _{ 30 }{ 15 } =\log _{ 30 }{ (\frac { 30 }{ 2 } ) } \\ \quad \quad \quad \quad =\quad \log _{ 30 }{ 30 } -\quad \log _{ 30 }{ 2 } =\quad 1\quad -\log _{ 30 }{ 2 } \quad \quad \quad \quad \\ \therefore \quad a\quad +\quad b\quad =\quad 1\quad -\log _{ 30 }{ 2 } \quad \\ =>\quad \log _{ 30 }{ 2 } =\quad 1-a-b\\ =>\quad 4\log _{ 30 }{ 2 } =\quad 4(1-a-b)\\ =>\quad \log _{ 30 }{ { 2 }^{ 4 } } =\quad 4(1-a-b)\\ =>\quad \log _{ 30 }{ 16 } =\quad 4(1-a-b)\quad \quad$

Moderator note:

Simple straightforward approach.

It is easier to motivate this solution by starting with $\log_{30} 16$ to explain why we're interested in $\log_{30} 2$ .

Gautam Jha
May 27, 2015

$\Large a+b=log_{30}3+log_{30}5=log_{30}15=1-log_{30}2$

$\implies \Large log_{30}2=1-a-b$

Also,

$\Large log_{30}16=log_{30}2^4=4log_{30}2$

$\implies \Large log_{30}16=4(1-a-b)$

Similar method over here! :)

Noel Lo - 6 years ago

Noel Lo
May 30, 2015

Hi Nihar, it was a nice problem!!!

JEE Novice - (23)

This question is a part of JEE Novices .

3 solutions

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