If $1$ is the solution of the equation ${log}_{20} x {log}_{21} x {log}_{22} x = {log}_{20} x {log}_{21} x + {log}_{21} x {log}_{22} x + {log}_{22} x {log}_{20} x$ the find the other solutions:

Open in App

Solution

$d i v i d e b y {log}_{20} x . {log}_{21} x . {log}_{22} x$
${log}_{x} 22 + {log}_{x} 20 + {log}_{x} 21 = 1$
${log}_{x} (22.20.21) = 1$
$x = 22.20.21 = 9240$

Suggest Corrections

Similar questions

x^{3} - 1 = 0

1

(s)

Find the value of $k,$ if $x = 2, y = 1$ is a solution of the equation $2 x + 3 y = k$ . Find two more solutions of the resultant equation.

3 x^{2} = 8 x + 2 k + 1

k

y = k e^{x}

(0, 1)

2^{| y |} - ∣ ∣ 2^{y - 1} - 1 ∣ ∣ = 2^{y - 1} + 1

Join BYJU'S Learning Program