Easy! Just multiply by 2!

Algebra Level 3

$\large \log_8 16 + \log_{16} 8= \ ?$

2 solutions

Mehul Arora
Jun 16, 2015

$log_{{2}^{3}} {2}^{4}+ log_ {{2}^{4}} {2}^{3}$

$\dfrac {4}{3} \times 1+ \dfrac {3}{4} \times 1$ ( $log_x x=1$ )

$\dfrac{4}{3}+ \dfrac {3}{4}$

$\dfrac {25}{12}$

$=2.083$

For proper understanding base change rule - $log_a{b} = \frac{logb}{loga}$

sandeep Rathod - 5 years, 12 months ago

Hi, what I was wondering was why is it written " Decimal OK " in the answer box.

Abhijeet Verma - 5 years, 12 months ago

Decimal OK means that the Problem poster has given the answer to be in Decimals. So, You have to answer in Decimals too :D

Mehul Arora - 5 years, 12 months ago

Yeah, but why OK and that too in caps :D

Abhijeet Verma - 5 years, 12 months ago

FIITJEE ones! @Mehul Arora

Anik Mandal - 5 years, 12 months ago

Yeah, I know! Easy, weren't they? :P

Mehul Arora - 5 years, 12 months ago

$\frac { 25 } { 12 }$ is not equal to $2.083$ .

The exact value of it is $2.083333333333 \cdots$ , so it is not $2.083$ .

. . - 3 months, 3 weeks ago

Towhidd Towhidd
Jul 26, 2015

say, log8(16)=x or, 8^x=16 or, (2^3)^x=2^4 or, 2^(3x)=2^4 or, 3x=4 or, x=4/3

again, log16(8)=y or, 16^y=8 or, (2^4)^y=2^3 or, 2^4y=2^3 or, 4y=3 or, y=3/4

now, x+y=(4/3)+(3/4)=(16+9)/12=25/12=2.083