Question

How to solve `log _{x}36+log _{18}3x = 3` ?

`log _{x}36+log _{18}3x = 3`

Answer 1

`x in {324; 6}`

Explanation:

`frac{log_2 36}{log_2 x} + frac{log_2 3 + log_2 x}{log_2 18} = 3`

`log_2 x = y`

`log_2 3 = K`

`frac{log_2 18 log_2 36}{y} + K + y = 3 log_2 18`

`frac{log_2 18 log_2 36}{y} + y = 3 (1 + 2 K) - K`

`c + y^2 = - by`

`b = - 3 - 5K`

`c = (1 + 2 K)(2 + 2 K) = 4K^2 + 6K + 2`

`Delta = (5K + 3)^2 - 16K^2 - 24K - 8`

`Delta = 9K^2 + 6K + 1`

`y = log_2 x = (3 + 5K pm (3K + 1))/2`

`x_1 =2 wedge (2 + 4K) = 4 * 3^4`

`x_2 =2 wedge (1 + K) = 2 * 3`

Answer 2

`color(blue)(x=324, x=6)`

Explanation:

`log_x36=log_18(36)/(log_18(x)` (change of base)

`log_18(36)/(log_18(x))+log_18(3x)=3`

`(2log_18(6))/(log_18(x))+log_18(x)+log_18(3)=3`

Let `m=log_18(x)`

`(2log_18(6))/m+m+log_18(3)=3`

`(2log_18(6))+m^2+log_18(3)m=3m`

`m^2+m(log_18(3)-3)+2log_18(6)=0`

Using quadratic formula:

`m=(-(log_18(3)-3)+-sqrt(((log_18(3)-3))^2-4(2log_18(6))))/2`

`m=(-(log_18(3)-3)+ sqrt(((log_18(3)-3))^2-(8log_18(6))))/2`

`=2`

`m=(-(log_18(3)-3)-sqrt(((log_18(3)-3))^2-(8log_18(6))))/2`

`=0.6199062340`

`m=log_18(x)`

`log_18(x)=2`

`18^(log_18(x))=18^2`

`x=18^2=324`

`log_18(x)=0.6199062340`

`18^(log_18(x))=18^(0.6199062340)`

`x=18^(0.6199062340)=6`