But All The Numbers Are Different!

Algebra Level 2

$a = \log_{10} 2 , b = \log_{10} 3 .$

Find the value of $\log_5 12$ in terms of $a$ and $b$ .

\displaystyle \cfrac { 2a+b }{ 1-a }

\displaystyle \cfrac { a+2b }{ 1+a }

Insufficient information None of these choices

1 solution

Soumo Mukherjee
Dec 20, 2014

Let $\displaystyle \log _{ 5 }{ 12 } =x$ .

Then $\displaystyle { 5 }^{ x }=12$ . Taking $\displaystyle \log _{ 10 }{ }$ at both sides we get: $\displaystyle \log _{ 10 }{ { 5 }^{ x } } =\log _{ 10 }{ 12 }$ or $\displaystyle x\log _{ 10 }{ { 5 } } =\log _{ 10 }{ 12 }$ .

Therefore. $x=\cfrac { \log _{ 10 }{ 12 } }{ \log _{ 10 }{ { 5 } } } =\cfrac { \log _{ 10 }{ \left( 4\times 3 \right) } }{ \log _{ 10 }{ { \left( { 10 }/{ 2 } \right) } } } =\cfrac { 2\log _{ 10 }{ 2 } +\log _{ 10 }{ 3 } }{ \log _{ 10 }{ { 10 } } -\log _{ 10 }{ 2 } } =\cfrac { 2a+b }{ 1-a }$

super solution can you say me some tips for doing this type of problems please

Sai Reddy - 6 years, 5 months ago

thanx but still 3 votes only!! :p

what about logarithm problems? most of the log problem demands its application of property which are not very difficult. Actually logarithm is inverse of exponential function. So if you could handle exp func pretty well then log problems becomes easier.

Soumo Mukherjee - 6 years, 5 months ago

But All The Numbers Are Different!

1 solution

0 pending reports