Question

Help? logs.

Answer 1

what logs?

Answer 2

yea what logs

Answer 3

http://web2.0calc.com/ Rose for then your doing logarithmic form try useing that calc

Answer 4

logarithmic form Is a other type of math equations

Answer 5

or logs

Answer 6

\[\log_{9}(-11x+2)=\log_{9}(x^{2}+30)\]
Do I divide by \[\log_{9} \] on both sides?
I need to find x

Answer 7

no you don't divide

Answer 8

you subtract

Answer 9

No I divide because it's opposite of multiplying.

Answer 10

1- log(A-B)=logA/logB

Answer 11

yes you got it

Answer 12

In Logs, we don't say like that..

Answer 13

2- bring log x in one side

Answer 14

we say, "take the antilogarithm" on both sides..

Answer 15

It is simple, don't make it hard..

Answer 16

@AJ01 you're not making sense
@waterineyes then how do I do it?

Answer 17

you have log base 9, on both the sides, so take antilog:
\[If \quad \ln_a(x) = \ln_a(y), \ \ then \ \ x = y\]

Answer 18

So:
\(\log_9(-11x+2) = \log_9(x^2 +30)\)

\(\implies -11x + 2 = x^2 + 30\)

Answer 19

Now, this is just a quadratic equation, find \(x\)..

Answer 20

@AJ01 Also this is correct formula:

\[\log(a) - \log(b) = \log(\frac{a}{b})\]

Answer 21

So I use the quadratic equation
\[x=\frac{ -b \pm \sqrt{b^{2}-4ac} }{ 2a }\]
\[x^{2}+11x+28\]

Answer 22

\[x=\frac{ -11 \pm \sqrt{11^{2}-4*1*28} }{ 2*1 }\]

Answer 23

you can also use Factorization Method if you want.. 7 and 4 are the factors..

Answer 24

\[x^2 + 11x + 28 = 0 \\ (x+7)(x+4) = 0 \\ x = -7 \quad or \ \quad x = -4\]

Answer 25

Getting??

Answer 26

Oh right! I always forget about factoring.

Answer 27

Good..

Answer 28

Could you help me with another. It's a little different.

Answer 29

Sure..