Question

log base 16 of 32 - log base 25 of 125 + log base 9 of 27​

Answer 1

$log_{16}32 - log_{25}125 + log_{9}27$
Using base change relation,
$\frac{ log_{2}32 }{ log_{2}16 } - \frac{ log_{5}125 }{ log_{5}25 } + \frac{ log_{3}9 }{ log_{3}27 }$

Solving this furthur,
$\frac{5}{4} - \frac{3}{ 2} + \frac{2}{3} \\ = \frac{15 - 18 + 8}{12} \\ = \frac{5}{12}$

Answer 2

log

16

32−log

25

125+log

9

27

Using base change relation,

\frac{ log_{2}32 }{ log_{2}16 } - \frac{ log_{5}125 }{ log_{5}25 } + \frac{ log_{3}9 }{ log_{3}27 }

log

2

16

log

2

32

−

log

5

25

log

5

125

+

log

3

27

log

3

9

Solving this furthur,

\begin{lgathered}\frac{5}{4} - \frac{3}{ 2} + \frac{2}{3} \\ = \frac{15 - 18 + 8}{12} \\ = \frac{5}{12}\end{lgathered}

4

5

−

2

3

+

3

2

=

12

15−18+8

=

12

5