If $log (a + b) = log a + log b$ , find $a$ in terms of $b$ .

a = \frac{a}{b - 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

a = \frac{b}{b + 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

a = \frac{b}{a - 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

a = \frac{b}{b - 1}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $a = \frac{b}{b - 1}$
$log (a + b) = log a + log b$
$∴ log (a + b) = log (a . b)$ ...(Property of logarithm)
$∴ a + b = a b$
$∴ a = \frac{b}{b - 1}$

Suggest Corrections

Similar questions

log (a + b) = log a + log b

a

b

[l o g_{b} a l o g_{c} a - l o g_{a} a] + [l o g_{a} b l o g_{c} b - l o g_{b} b] + [l o g_{a} c l o g_{b} c - l o g_{c} c] = 0

2^{(\sqrt{l o g_{a} \sqrt[4]{a b} + l o g_{b} \sqrt[4]{a b}} - \sqrt{l o g_{a} \sqrt[4]{b / a} + l o g_{b} \sqrt[4]{a / b}}) . \sqrt{l o g_{a} b}}

= {\begin{matrix} 2, b \geq a \geq 1 2^{l o g_{a} b}, 1 < b < a \end{matrix}

log \frac{a}{b} + log \frac{b}{a} = log (a + b),

[l o g_{b} a l o g_{c} a - l o g_{a} a] + [l o g_{a} b l o g_{c} b - l o g_{b} b] + [l o g_{a} c l o g_{b} c - l o g_{c} c] = 0, t h e n a b c =

Join BYJU'S Learning Program