3 years ago

Please help me outtttttttttt

Mathematics

1 answer:

Evgen [1.6K]3 years ago

3 0

Answer:

is correct

Step-by-step explanation:

1/12

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3 years ago

How to solve: <img src="https://tex.z-dn.net/?f=log_%7B3%7D%20%28x%2B6%29%20%2B%20log_%7B3%7D%28x-6%29%20%3D%204" id="TexFormula

Shalnov [3]

Collapse the logarithms into one:

$\log_3(x+6)+\log_3(x-6)=\log_3((x+6)(x-6))=4$

Then write both sides as powers of 3:

$3^{\log_3((x+6)(x-6))}=3^4$

$(x+6)(x-6)=81$

Simplify the left side and solve for $x$ :

$x^2-36=81$

$x^2=117$

$x=\pm\sqrt{117}=\pm3\sqrt{13}$

But we're not done yet. Notice that both $x+6$ and $x-6$ are negative when $x=-3\sqrt{13}$ . We can't take the logarithm of a negative number (the result isn't real-valued, anyway), so we throw this solution out, and we're left with just