How do you simplify #Log(x+4)=2log(x-2)#?

Question

Lincoln Anderson · Accepted Answer

#x=5#

Simplify the right hand side using the logarithm rule:

#color(white)(sss)alog_b(c)=log_b(a^c)#

#log(x+4)=log((x-2)^2)#

Exponentiate both sides, which undoes both logarithms, leaving us with:

#x+4=(x-2)^2#

Distribute and simplify.

#x+4=x^2-4x+4#

#0=x^2-5x#

#0=x(x-5)#

#x=0,5#

However, the answer #x=0# is thrown out, since plugging in #0# in #2log(x-2)# would result in having to find the logarithm of a negative number, which is impossible.

Thus, the only valid answer is

#x=5#

graph{log(x+4)-2log(x-2) [-8.24, 17.07, -3.01, 9.65]}

Ariana Alvarez · Answer

undefined To simplify the equation $ \log(x + 4) = 2\log(x - 2) $, we'll first use the properties of logarithms to condense the equation.

$$ \log(x + 4) = \log((x - 2)^2) $$

Now, according to the property $ \log_b(a^n) = n\log_b(a) $, we can rewrite $ \log((x - 2)^2) $ as $ 2\log(x - 2) $.

So, the simplified equation becomes:

$$ \log(x + 4) = \log((x - 2)^2) $$

Since both sides of the equation are now in terms of the same logarithm, we can drop the logarithms:

$$ x + 4 = (x - 2)^2 $$

Next, we expand $ (x - 2)^2 $:

$$ x + 4 = x^2 - 4x + 4 $$

Now, we'll move all terms to one side of the equation to solve for $ x $:

$$ x - x^2 + 4x - 4 = 0 $$

$$ -x^2 + 5x - 4 = 0 $$

Now, we can solve this quadratic equation for $ x $ using factoring, completing the square, or the quadratic formula. Once we find the solutions for $ x $, we can check if they satisfy the original equation.