Log problem

Note first that in general $\log_{a}b = \dfrac{1}{\log_{b}a}$ for $a, b \gt 0$ .

So the expression becomes

$\log_{3}(3*8) *\log_{3}72 - \log_{3}(3*72) *\log_{3}8 =$

$(1 + \log_{3}8)*\log_{3}72 - (1 + \log_{3}72)*\log_{3}8 =$

$\log_{3}72 - \log_{3}8 = \log_{3} \frac{72}{8} = \log_{3}9 = \boxed{2}$ .

loga b = 1/logb a

1.log3 24 x log 3 72 - log3 216 x log3 8

2.log3 (8x3) x log3 72 - log3 (8 x 3^3) x log3 8

3.(log3 8 + 1 ) x (log3 8 + 2) - (log3 8 + 3 ) (log3 8)

4.let log3 8 be x

5.(x+1) (x+2 ) - (x+3) (x)

6.x^2 +2x + x +2 - x^2 - 3x =2

convert everything in form of $(\log _{ 3 }{ 8 }$ )

the expression becomes

$[(\log _{ 3 }{ (8*3) } )*(\log _{ 3 }{ (8*{ 3 }^{ 2 }) } )]-[(\log _{ 3 }{ (8*{ 3 }^{ 3 }) } )*(\log _{ 3 }{ (8) } )]$

the simplified expression is

$[(\log _{ 3 }{ 8 } +1)*(\log _{ 3 }{ 8 } +2)]\quad -\quad [(\log _{ 3 }{ 8 } +3)*(\log _{ 3 }{ 8 } )]$

now take $(\log _{ 3 }{ 8 } ) = a$

the expression becomes

$\left( a+1 \right) \left( a+2 \right) -\left( a+3 \right) \left( a \right) )$

= ${ a }^{ 2 }+\left( 3*a \right) +2-{ a }^{ 2 }-\left( 3*a \right)$

= 2

Bearing in mind the identity Brian made reference to in his working, we can rewrite the expression to be:

$log_3 24 log_3 72 - log_3 216log_3 8 = log_3 24 (log_3 72) - (log_3 (72 \times 3))(log_3 \frac{24}{3})$

$= log_3 24 (log_3 72) - (log_3 72 +1)(log_3 24-1) = log_3 72 -log_3 24 + 1$

$= log_3 \frac{72}{24} + 1 = log_3 3 + 1 = 1+1 = \boxed{2}$

just use base change property , assume log2 base 3 to be equal to a x and then just simplify , then x will get cancelled and 2 will remain. So Ans=>2