Question

1/2 log2(x+4)-1/2 log2 (x-4)=2

Answer 1

\[1/2 \log _{2}(x+4)-1/2\log _{2}(x-4)=2\]

Answer 2

perform the power rule on your two logarithms first..what do you have?

Answer 3

@lgbasallote Is it 2 log2 1/2?

Answer 4

are you asking if that's the final answer? coz im thinking the final answer doesnt have a log

Answer 5

No. You said perform the power rule... on my first log this is the answer I got. Is this correct? Can you show me the steps please?

Answer 6

power rule states that

\(\Large a\log_{b} c = \log_{b} a^c\)

so

\(\Large \frac{1}{2} \log_{2} (x + 4) = \log_{2} (x+4)^{\frac{1}{2}}\)

you understand that right?

btw..if you need help on the properties do check this out http://openstudy.com/users/lgbasallote#/updates/4fa45f8fe4b029e9dc34e0b5

might help

Answer 7

1/2log2(x+4)−1/2log2(x−4)=2
log2(x+4)-log2(x-4)=4
log2(x+4/x-4)=4
x+4/x-4=2^4
Solve as needed.

Answer 8

\[\log_{2}\frac{\sqrt{x+4}}{\sqrt{x-4}}=2 \]

Answer 9

Put it in exponential form.

Answer 10

oh it's still log form sorry :P

Answer 11

\[\frac{\sqrt{x+4}}{\sqrt{x-4}}=4\]

Answer 12

Square both sides.

Answer 13

\[\frac{x+4}{x-4}=16\]

Answer 14

\[x+4=16x-64\]

Answer 15

Finish it up.

Answer 16

How did you get 16? I thought it was x+4/x-4 = 2^2 which would be x=4/x-4 = 4.

Answer 17

\[\frac{\sqrt{x+4}}{\sqrt{x-4}}=2^2\]

Answer 18

When you square both sides, it eliminates the radicals on the left and squares the 4 on the right. 4 squared is 16

Answer 19

So then I will get x=4+16x-64 correct?

Answer 20

@Mertsj and @lgbasallote is my answer 4.5333=x?

Answer 21

Whatever 68/15 is

Answer 22

So when I got to 68/15 I am correct at that point as well, correct?

Answer 23

Yep

Answer 24

Thank you

Answer 25

yw