Is the information sufficient 7?

Algebra Level 3

If log 12 36 = x \log_{12} 36 = x , then log 12 48 = a x + b \log_{12} 48 = ax + b , where a , b a, b are integers. Find the value of

( b a ) b + a \large (b - a)^{b + a}


For more problems like this, try answering this set .


The answer is 16.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Ka Sigmund
Jun 29, 2017

l o g 12 48 = a l o g 12 36 + b = t log_{12}{48}=a log_{12}{36}+b=t

1 2 t = 48 = ( 3 6 a ) ( 1 2 b ) 12^t=48=(36^a)(12^b)

2 4 3 1 = 2 2 a + 2 b 3 2 a + b 2^43^1=2^{2a+2b}3^{2a+b}

gives, b=4-1=3 and a= 4 6 2 \frac{4-6}{2} =(-1).

therefore, ( a b ) a + b = 4 2 = 16 (a-b)^{a+b}=4^2=16

Ravneet Singh
Jun 29, 2017

log 12 36 + log 12 48 = log 12 ( 36 × 48 ) = log 12 1728 = log 12 1 2 3 = 3 \log_{12} 36 + \log_{12} 48 = \log_{12} (36 \times 48) = \log_{12} 1728 = \log_{12} 12^3 = 3

log 12 48 = 3 log 12 36 \log_{12} 48 = 3 - \log_{12} 36

log 12 48 = 3 x = ( 1 ) x + 3 \log_{12} 48 = 3 - x = (-1)x + 3

on comparing we get a = 1 , b = 3 a = -1, b = 3

( b a ) b + a = 4 2 = 16 \implies \large (b - a)^{b + a} = 4^2 = \boxed{16}

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...