Question

Solve this log question
If a2 + b2 = 7ab, show that log (a+b)/3 = 1/2(log a+log b)​

Answer 1

Gɪᴠᴇɴ :-

• a² + b² = 7ab

Tᴏ SHᴏw :-

• log (a+b)/3 = 1/2(log a+log b) ?

Sᴏʟᴜᴛɪᴏɴ :-

→ a² + b² = 7ab

Adding 2ab both sides,

→ a² + b² + 2ab = 7ab + 2ab

→ (a + b)² = 9ab

Square - Root Both sides,

→ (a + b) = 3(ab)^(1/2)

Taking log both sides now,

→ log(a + b) = log[ 3(ab)^(1/2)]

using log(x * y) = log(x) + log(y) in LHS now,

→ log(a + b) = log(3) + log(ab)^(1/2)

→ log(a + b) - log(3) = log(ab)^(1/2)

using log(x) - log(y) = log(x / y) in RHS Now,

→ log(a + b / 3) = log(ab)^(1/2)

→ log(a + b / 3) = (1/2)*log(ab)

Again using using log(x * y) = log(x) + log(y) in LHS now,

→ log (a+b)/3 = 1/2(log a+log b) (Hence, Proved).

Answer 2

$\rule{200}3$

$\huge\tt{GIVEN:}$

• a² + b² = 7ab

$\rule{200}3$

$\huge\tt{TO~SHOW:}$

• log (a+b)/3 = 1/2(log a+log b)

$\rule{200}3$

$\huge\tt{SOLUTION:}$

↪a² + b² = 7ab

↪a² + b² + 2ab = 7ab + 2ab {Adding 2ab to both sides}

↪(a + b)² = 9ab

↪(a + b) = 3(ab)^½ {squaring both roots}

↪log(a+b) = log[3 (ab)^½] {Taking log both sides}

↪log (a+b) - log 3 = log(ab)^½

↪log(a+b/3) = log(ab^½)

↪log(a+b/3) = ½ × log(ab)

↪log (a+b)/3 = ½(log a + log b)

✓ Proved

$\rule{200}3$