Question

solve for x? logbase2(5-x)-logbase2(2+x^2)=1?

Answer 1

but...howd you get there?

Answer 2

\[\log_{2} (5-x)-\log_{2}(2+x^2)=1\]\[\log_{2} (5-x)/(2+x^2)=\log_{2}2\]\[(5-x)/(2+x^2)=2\]\[5-x=4+2x^2\]\[2x^2+x-1=0\]
\[Δ=1^2-4\times(-1)2=9\]\[x_1,_2=(-1\pm \sqrt{9})/2\times2\]
\[x=1/2\]or\[x=-1\]

Answer 3

excellent work...........i just guessed :)

Answer 4

Hi Christal,

The key in what Johnny wrote is recognizing that the right side of the equation can be rewritten:

\[\log_{2} 2 = 1\]

After that, every term in your equation has log base 2... so then you can effectively eliminate them after some grouping. remember the log rules earlier?

Starting with:

\[\log_{2} (5-x)-\log_{2}(2+x^{2}) = 1 = \log_2{2}\]

Remembering that (log a - log b = log a/b), we can get
\[\log_{2} \frac{5-x}{2+x^{2}}=\log_{2}2\]

Now, since they are all log base 2, get rid of that stupid log!
\[\frac{(5-2)}{2+x^{2}}=2\]

From there simplify and solve!

Answer 5

wow thanks that really helped!

Answer 6

could uu help me with another?

Answer 7

Hi Kmousou. I just learned today.

if you have the fraction x/y, in the equation editor just type frac{x}{y}

Answer 8

Sorry, site seems really slow for me, Christal. Sure, I will help if I can.

Answer 9

okay. h ow about ln(2x+7)=-1?

Answer 10

Give you a hint first, remember:

\[\ln e^{x} = x\]

So, your right side equation (the number -1) can be written as:

\[\ln e^{-1}\]

From there you can effectively get rid of your ln. Make sense so far? So what would you do next?

Answer 11

ln2x times ln7= lne^-1..so get rid of ln?

Answer 12

Actually, the left hand side is fine already. It is in a good form for solving the equation. The goal is to have a single ln(of some values) = ln(of some values).

So,
\[\ln (2x+7) = \ln e^{-1}\]
then,
\[2x+7 = e^{-1} = \frac{1}{e}\]
Then solve for x.

Answer 13

Sorry,
\[\ln e^{1}\]
Should be
\[\ln e^{-1}\]

Answer 14

lol... should read (ln of e to the -1)... not sure why the -1 isn't showing up.

Answer 15

its alright i got it! yayyy! except the next one...

Answer 16

i see the negative one

Answer 17

!! cool.

Answer 18

[ln(x)]^2-2ln(x)=0. is there a difference?

Answer 19

So, if I understand what you wrote:
\[(\ln x)^{2} - 2 \ln {x} = 0\]

Remember, the goal is to put a ln(something) = ln (something), so
\[(\ln x)^{2} = 2 \ln {x}\]

Ahhgg, going to run out of time to do this one right, but

ln(x) * ln(x) = ln(x+x) (from the rules!) = ln(2x)

Look up the power rule for logs because 2 * ln(x) can be rewritten too. Once you have done that, you have the form you need to simplify and drop the ln.

Answer 20

thaaaank youu!! so much

Answer 21

you're welcome. See you later.