CameraIcon

SearchIcon

MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Definition of a Determinant

Show that: ...

Question

Show that: $2^{(\sqrt{l o g_{a} \sqrt[4]{a b} + l o g_{b} \sqrt[4]{a b}} - \sqrt{l o g_{a} \sqrt[4]{b / a} + l o g_{b} \sqrt[4]{a / b}}) . \sqrt{l o g_{a} b}}$
$= {\begin{matrix} 2, b \geq a \geq 1 2^{l o g_{a} b}, 1 < b < a \end{matrix}$

Open in App

Solution

Since $\sqrt{l o g_{a} \sqrt[4]{a b} + l o g_{b} \sqrt[4]{a b}}$
$= \sqrt{\frac{1}{4} l o g_{a} (a b) + \frac{1}{4} l o g_{b} (a b)}$
$= \sqrt{\frac{1}{4} (1 + l o g_{a} b + l o g_{b} a + 1} = \sqrt{⎛ ⎜ ⎝ \frac{l o g_{a} b + \frac{1}{l o g_{a} b} + 2}{4} ⎞ ⎟ ⎠}$
$= \sqrt{{⎛ ⎜ ⎜ ⎝ \frac{\sqrt{| l o g_{a} b |} + \frac{1}{\sqrt{| l o g_{a} b |}}}{2} ⎞ ⎟ ⎟ ⎠}^{2}}$
$= \frac{1}{2} (\sqrt{| l o g_{a} b |} + \frac{1}{\sqrt{| l o g_{a} b |}})$
Consider $\sqrt{l o g_{a} \sqrt[4]{(b / a)} + l o g_{b} \sqrt[4]{(a / b)}}$
$= \sqrt{\frac{1}{4} l o g_{a} (\frac{b}{a}) + \frac{1}{4} l o g_{b} (\frac{a}{b})}$
$= \sqrt{\frac{1}{4} (l o g_{a} b - 1 + l o g_{b} a - 1)} = \sqrt{\frac{l o g_{b} b + \frac{1}{l o g_{a} b} - 2}{4}}$
$= \frac{∣ ∣ ∣ \sqrt{| l o g_{a} b | - \frac{1}{\sqrt{| l o g_{a} b |}}} ∣ ∣ ∣}{2}$
Now, $P (say) = \sqrt{l o g_{a} \sqrt[4]{a b} + l o g_{b} \sqrt[4]{a b}} - \sqrt{l o g_{a} \sqrt[4]{b / a} + l o g_{b} \sqrt[4]{a / b}}$
$P = \frac{1}{2} ⎧ ⎨ ⎩ \sqrt{| l o g_{a} b | + \frac{1}{\sqrt{| l o g_{a} b |}}} + ∣ ∣ ∣ ∣ \sqrt{| l o g_{a} b |} - \frac{1}{\sqrt{| l o g_{a} b |}} ∣ ∣ ∣ ∣ ⎫ ⎬ ⎭$
Case I : If $b \geq a \geq 1, then$
$P = \frac{1}{2} {\sqrt{| l o g_{a} b |} + \frac{1}{\sqrt{| l o g_{a} b |}} - \sqrt{| l o g_{a} b |} + \frac{1}{\sqrt{| l o g_{a} b |}}}$

$= \frac{1}{\sqrt{l o g_{a} b}}$
$∴ 2^{P \sqrt{l o g_{a} b}} = 2^{1} = 2$
Case II : If 1 < b < a, then
$P = \frac{1}{2} {\sqrt{| l o g_{a} b |} + \frac{1}{\sqrt{| l o g_{a} b |}} + \sqrt{| l o g_{a} b |} - \frac{1}{\sqrt{| l o g_{a} b |}}}$
$= \sqrt{l o g_{a} b}$
$∴ 2^{P \sqrt{l o g_{a} b}} = 2^{l o g_{a} b}$

Suggest Corrections

thumbs-up

1

Similar questions

log (\frac{a + b}{4}) = \frac{1}{2} [log a + log b]

(a - b)^{2} = 12 a b

log (\frac{a + b}{2}) = \frac{1}{2} (log a + log b)

, then show that

a = b

Q. If a,b,c are distinct positive numbers each different from 1 such that

[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

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

a^{2} - 12 a b + 4 b^{2} = 0

log (a + 2 b) = \frac{1}{2} (log a + log b) + 2 log 2

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Introduction

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Definition of a Determinant

Standard XII Mathematics

Join BYJU'S Learning Program

CrossIcon