Question

Question with log, solve for x:

(log2X+logX)/(log2X-logX) = 3

(base of log is the same)

Answer 1

\[\frac{ \log2x+logx }{ \log2x-logx } = 3\]

Answer 2

show me the steps please

Answer 3

X=2

Answer 4

Sorry .

Answer 5

yes x=2

Answer 6

how?

Answer 7

log2x +logx = log2x^2

Answer 8

From the rules of logs
\[\log_{} 2x+\log_{} x=\log_{} 2x ^{2}\]
Do you understand this step?

Answer 9

that's how i translated it for the numerator yes

Answer 10

for the denominator it's log2x/x

Answer 11

so it's log2

Answer 12

So othe denominator is log 2. Correct.

Answer 13

and then???

Answer 14

So we now have by cross multiplying
\[\log_{} 2x ^{2}=3\log_{} 2\]
Understood?

Answer 15

oh i see!

Answer 16

so what i did was

Answer 17

and
\[3\log_{} 2=\log_{} 2^{3}=\log_{} 8\]

Answer 18

log2x^2/log2 , i just cancelled out the 2's on top and bottom, leaving logx^2=3, so that's wrong

Answer 19

yes i understand what you did there it makes sense, i just want to understand why i'm wrong

Answer 20

Yes, that was wrong.
\[\log_{} 2x ^{2}=\log_{} 8\]
So\[2x ^{2}=8\]

Answer 21

i just have one quick question, when i see this \[\frac{ \log2x ^{2} }{ \log2? }\] is it equivalent to log2x^2-log2 ?

Answer 22

(take out the question mark beside log2)

Answer 23

To understand where you went wrong, please look at this example:
\[\frac{\log_{} (2 \times 100)}{\log_{} 2}=\frac{\log_{} 200}{\log_{} 2}=7.644\]
If you try your 'cancelling' you would get a result of 2, which is obviously wrong.
Does this help you?

Answer 24

hmm interesting. it makes more sense now

Answer 25

yeah it's really more clear when you see it with actual numbers thank you so much.

Answer 26

The answer to your quick question is no, the reason being that both terms are already logarithms. Therefore you do not apply the rule that you would apply to a quotient of real numbers.

Answer 27

thank you sir i'll try to chew on it more, it was very helpful.

Answer 28

I'm glad to be of help :)