Question

log(2x-5) + log(x-6) = log(6x+20) what is the value of x?

Answer 1

Answer:

should help u little bit

Answer 2

Answer:

The value of x is:

$\dfrac{23\±\sqrt {449}}{4}$

Step-by-step explanation:

Given equation is:

$log(2x-5)+log(x-6)=log(6x+20)$

Here we are applying the formula of $log(a)+log(b)$.

$log(a)+log(b)=log(ab)$

Hence:

$log[(2x-5)(x-6)]=log(6x+20)$

Cancelling out the log from both side.

$(2x-5)(x-6)=(6x+20)\\2x^2-12x-5x+30=6x+20\\2x^2-17x+30=6x+20\\2x^2-17x-6x+30-20=0\\2x^2-23x+10=0$

Here we are applying sridharacharya formula:

$x=\dfrac{-b\±\sqrt {b^2-4ac}}{2a}$

Values are:

$a=2\\b=-23\\c=10$

$x=\dfrac{23\±\sqrt {(23)^2-4\cdot 2\cdot 10}}{2\cdot 2}\\x=\dfrac{23\±\sqrt {529-80}}{4}\\x=\dfrac{23\±\sqrt {449}}{4}$