Solve: log base 4 of (x – 6) + log base 4 of x = 2?

log base 4 of (x – 6) + log base 4 of x = 2
=>log base 4 of x(x – 6) = log base 4 of 4^2
=> x(x-6) = 16
=> x^2-6x-16 = (x-8)(x+2) = 0
x = 8, -2(rejected)
Answer: x = 8
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Attn: The domain of log x is x > 0.

log base 4 of x(x-6) = 2
4^2 = x^2 - 6x
x^2 - 6x -16 = 0
(x-8)(x+2) = 0
x = 8, -2

plug them into the original, the -2 doesn't work, so x=8 is your answer,

use log rule
that says loga +logb=log(ab)
iff the logs have the same base.
then you get
log base 4 (x(x-6))=2

Property of logs
if log base a of b=c
then a^c=b

in your case 4^2=x(x-6)
Simplifying we get
x^2-6x-16=0
(x-8)(x+2)=0
so x=8 or x=-2
NOT FINISHED YET!
Plug both in to see if they work.
x=-2 will NOT work since log base 4 of -2
is UNDEFINED DUE TO THE FACT THE IMPUTS
OF A LOG HAVE TO BE STRICTLY GREATER THAN
0
enjoy

log(base 2) (2x-4) = 2 + log(base 2) (x^2-6) => log(base 2) (2x-4) - log(base 2) (x^2-6) = 2 => log(base 2) [(2x - 4) / (x^2 - 6) ] = 2 => (2x - 4) /(x^2 - 6) = 2^2 = 4 => 2x - 4 = 4x^2 - 24 => divide by 2 x - 2 = 2x^2 - 12 => 2x^2 - x - 10 = 0 => 2x^2 - 5x + 4x - 10 = 0 => x(2x - 5) + 2(2x - 5) = 0 => (2x - 5)(x + 2) = 0 since log of negative numbers is not defined 2x = 5 x = 5/2

log_4(x - 6) + log_4(x) = 2
log_4[x(x - 6)] = 2
x(x - 6) = 4^2
x^2 - 6x = 16
x^2 - 6x - 16 = 0
x^2 + 2x - 8x - 16 = 0
(x^2 + 2x) - (8x + 16) = 0
x(x + 2) - 8(x + 2) = 0
(x + 2)(x - 8) = 0

x + 2 = 0
x = -2

x - 8 = 0
x = 8
(BTW, x cannot be negative.)

∴ x = 8

Let log be log to base 4 :-

log ( x - 6 ) + log x = 2

log [ x ( x - 6 ) ] = 2

x ( x - 6 ) = 16

x^2 - 6x - 16 = 0

( x - 8 ) ( x + 2 ) = 0

x = 8 is acceptable

log base4 x^2-6x = 2
then you get 4squared = xsquared - 6x
then you get xsquared - 6x - 16 = 0 and factor that to solve for x