First slide

Trigonometric Identities

Question

Consider equations $x^{\log_{y} x} = 2 and y^{\log_{x} y} = 16$

Moderate

Question

The value of x is

A

$2^{\sqrt[3]{2}}$
B

$2^{\sqrt[3]{4}}$
C

$2^{\sqrt[3]{64}}$
D

$2^{\sqrt[3]{256}}$

Solution

$\begin{aligned} let \log_{y} x = t \\ then x = y^{t} - - - (1) \end{aligned}$
$\begin{array}{l} now, x^{\log_{y} x} = 2 becomes \\ x^{t} = 2 \\ x = 2^{1 / t} - - - - - (2) \end{array}$
$\begin{array}{l} and y^{\log_{x} y} = 16 becomes \\ y^{1 / t} = 2^{4} \\ y = 2^{4 t} - - - - - - (3) \end{array}$
Putting the values of x and y in (1), we get
$\begin{array}{ll} 2^{1 / t} = 2^{4 t^{2}} \\ \Rightarrow 4 t^{3} = 1 \\ ∴ t = {(\frac{1}{4})}^{1 / 3} - - - - (4) \end{array}$
Using (4) and (2), we get $x = (2)^{(4)^{1 / 3}} = 2^{\sqrt[3]{4}}$
Using (4) and (3), we get $y = (2)^{(4)^{2 / 3}} = 2^{\sqrt[3]{16}}$

Question

The value of y is

A

$2^{\sqrt[3]{2}}$
B

$2^{\sqrt[3]{4}}$
C

$2^{\sqrt[3]{64}}$
D

$2^{\sqrt[3]{256}}$

Solution

$\begin{aligned} let \log_{y} x = t \\ then x = y^{t} - - - (1) \end{aligned}$
$\begin{array}{l} now, x^{\log_{y} x} = 2 becomes \\ x^{t} = 2 \\ x = 2^{1 / t} - - - - - (2) \end{array}$
$\begin{array}{l} and y^{\log_{x} y} = 16 becomes \\ y^{1 / t} = 2^{4} \\ y = 2^{4 t} - - - - - - (3) \end{array}$
Putting the values of x and y in (1), we get
$\begin{array}{ll} 2^{1 / t} = 2^{4 t^{2}} \\ \Rightarrow 4 t^{3} = 1 \\ ∴ t = {(\frac{1}{4})}^{1 / 3} - - - - (4) \end{array}$
Using (4) and (2), we get $x = (2)^{(4)^{1 / 3}} = 2^{\sqrt[3]{4}}$
Using (4) and (3), we get $y = (2)^{(4)^{2 / 3}} = 2^{\sqrt[3]{16}}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App