Question

Log4(x)+log4(x-3)=log4(-7x+21)

Answer 1

Answer:

$x_{1}=-7 \\x_{2}=3$

Step-by-step explanation:

The given equation is

$log_{4} (x)+log_{4}(x-3)=log_{4}(-7x+21)$

First, we need to use the product property of logarithms

$log_{4}(x)+log_{4}(x-3)=log_{4}(x(x-3))$

Replacing this in the equation, er have

$log_{4}(x(x-3))=log_{4}(-7x+21)$

Now we can cancel logarithms

$x(x-3)=-7x+21$

Then, we use the distributive property and solve for $x$

$x^{2} -3x=-7x+21\\x^{2} -3x+7x-21=0\\x^{2} +4x-21=0$

We need to find two number which product is 21 and which difference is 4. Those numbers are 7 and 3, because 7x3 = 21 and 7 - 3 = 4.

$x^{2} +4x-21=(x+7)(x-3)$

There are two solutions

$x_{1}=-7 \\x_{2}=3$

Answer 2

Log4(x(x-3))=Log4(-7x+21)

Log4(X the power of two -3x)=Log4(-7x21)

X the power of two -3x=-7x+21

x=3,x=-7.