# Solve: log(2) x – log(x) 8 = 2

justaguide | College Teacher | (Level 2) Distinguished Educator

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We have to solve log(2) x – log(x) 8 = 2

log(2) x – log(x) 8 = 2

use the property of logarithm: log a^b = b*log a

=> log(2) x – 3*log(x) 2 = 2

Use the property of logarithm: log(a)b  = 1/log(b) a

=> log(2) x - 3/log(2) x = 2

=> (log(2) x)^2  - 2*log(2) x - 3 = 0

=> (log(2) x)^2  - 3*log(2) x + log(2) x - 3 = 0

=> (log(2) x)[log(2) x – 3] + 1[log(2) x – 3] = 0

=> [log(2) x  + 1][ log(2) x – 3] = 0

=> log(2) x = -1 and log(2) x = 3

=> x = 1/2  and  x = 2^3 = 8

The solution of the equation is x = 1/2 and x = 8

tonys538 | Student, Undergraduate | (Level 1) Valedictorian

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The equation `log_2 x - log_x 8 = 2` has to be solved.

For logarithms the following relations hold `log_b a = 1/(log_a b)` , `log a^b = b*log a`

`log_2 x - log_x 8 = 2`

`log_2 x - log_x 2^3 = 2`

`1/(log_x 2) - 3*log_x 2 = 2`

`1 - 3*(log_x 2)^2 = 2*(log_x 2)`

`3*(log_x 2)^2 + 2*(log_x 2) - 1 = 0`

`3*(log_x 2)^2 + 3*(log_x 2) - (log_x 2) - 1 = 0`

`3*log_x 2(log_x 2 + 1) - 1(log_x 2 + 1) = 0`

`(3*log_x 2 - 1)(log_x 2 + 1) = 0`

`3*log_x 2 - 1 = 0`

`log_x 2 = 1/3`

`log_2 x = 3`

`x = 2^-3 = 8`

`log_x 2 + 1 = 0`

`log_x 2 = -1`

`1/(log_2 x) = -1`

`x = 2^-1 = 1/2`

The solution of the equation `log(2) x - log(x) 8 = 2` is x = 1/2 and x = 8

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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We'll write log(x) 8 = 1/log (8) x

log(2) x =log(8) x*log(2) 8

log(2) 8= log(2) (2^3)

We'll use the power property of logarithms:

log(2) (2^3) = 3log(2) (2) = 3

log(2) x =log(8) x*3 => log(8) x=log(2) x/3

The equation will become:

log(2) x - 3/log(2) x - 2 = 0

[log(2) x]^2 - 2log(2) x - 3 = 0

We'll replace log(2) x by t:

t^2 - 2t - 3 = 0

t1 = 3, t2 = -1

log(2) x = 3 => x = 2^3 => x = 8

log(2) x = -1 => x = 2^-1 => x = 1/2

The solutions of the equation are: {1/2 ; 8}.