Please help I've been have trouble with it.

Mathematics

2 answers:

sashaice [31]2 years ago

8 0

Answer:

Step-by-step explanation:

kakasveta [241]2 years ago

3 0

It should be Answer D. I hope this helps. I also hope you have a great day!!!

You might be interested in

Can someone plz help me with Algebra 2....... :) Will give Brainliest!!

Anarel [89]

Using logarithms property of log(x)+log(y)=log(xy)
so here, you can sum the equation to;
$log((x+6)*(x-6))=2$
so you can simply say that;
$log_{8}((x+6)( x-6))=2$
and by multiplying (x+6)*(x-6)
$log_{8}(x^2-36)=2$
and as you know also that;
$a^{b}=c$ is same as $log _{a}c=b$
so you can simply state it as;
$8^2=x^2-36
64=x^2-36
64+36=x^2
x^2=100
x=10$
And you can check your work by substituting with 10 instead of x in the original function.
Hope this helps!

6 0

3 years ago

Solve for x in the equation x²-12x+59=0

lyudmila [28]

${x}^{2} - 12x + 59 = 0 \\ {x}^{2} - 12x + 36 + 23 = 0 \\ {x}^{2} - 12x + 36 = - 23 \\ {(x - 6)}^{2} = - 23 \\ x - 6 = + - \sqrt{ - 23} = + - i \sqrt{23} \\ x = 6 + - i \sqrt{23}$