Question

prove that 7log16/15+5log25/24+3log81/80=log2​

Answer 1

Log basis

• A product becomes a sum.
• A division becomes a subtraction.

$7\log\dfrac{16}{15} +5\log\dfrac{25}{24} +3\log\dfrac{81}{80}$

1. $\log\dfrac{16}{15}$ as $\log16 - \log15$
2. $\log\dfrac{25}{24}$ as $\log25 - \log24$
3. $\log\dfrac{81}{80}$ as $\log81 - \log80$

Let $a = \log2$, $b = \log3$, and $c = \log5$.

$7( \log16 - \log15 ) + 5( \log25 - \log24 ) + 3( \log81 - \log80 )$

$= 7 ( 4 \log2 - \log3 - \log5 ) + 5 ( 2 \log5 - 3\log2 - \log3 ) + 3 ( 4\log3 - 4\log2 - \log5 )$

$= 7 ( 4a - b - c ) + 5 ( 2c - 3a - b ) + 3 ( 4b - 4a - c )$

$= ( 28 - 15 - 12 ) a + ( - 7 - 5 + 12 ) b + ( - 7 + 10 - 3 ) c$

$=a$

$\therefore 7\log\dfrac{16}{15} +5\log\dfrac{25}{24} +3\log\dfrac{81}{80}=\log2$

Answer 2

Question= prove that 7log16/15+5log25/24+3log81/80=log2

Solution⬇️

The given expression E is,

E = 7 log (16/15) + 5 log (25/24) + 3 log (81/80)

= log (16/15)⁷ + log (25/24)⁵ + log (81/80)³

= log [ (16/15)⁷ × (25/24)⁵ × (81/80)³] —(1)

Now

(16/15)⁷ = (2⁴/3¹.5¹)⁷ = (2²⁸/3⁷.5⁷)

(25/24)⁵ = (5²/2³.3¹)⁵ = (5¹⁰/2¹⁵. 3³)

(81/80)³ = (3⁴/2⁴. 5¹)³ = (3¹²/2¹².5³)

So

[ (16/15)⁷ × (25/24)⁵ × (81/80)³]

= [{2²⁸ . 3¹² . 5¹⁰}/{2²⁷ . 3¹² . 5¹⁰}] = 2

Therefore,

E = log 2.