Question

log(x-4)+ log(x-3) = 2
(solve round the answer to 4 decimal places)

Answer 1

You can simplify that addition by using a log rule: http://www.purplemath.com/modules/logrules.htm

Answer 2

\[\large \log(a)+\log(b)=\log(a*b)\]

Answer 3

\[\large \log(x-4)+ \log(x-3) = 2 \\ \large \log((x-4)(x-3))=2\]

Answer 4

Let's suppose we have a natural log here
\[\large \log_e{(x-4)(x-3)}=2\\\large e^2=(x-4)(x-3)\\\large e^2=x^2-7x+12\\\large 0=x^2-7x+12-e^2\]
\[\large x=\frac{1}{2}(7-\sqrt{1+4e^2})=0.7361157\\ \large x=\frac{1}{2}(7+\sqrt{1+4e^2})=6.2638842\]

Answer 5

@Calcmathlete Can you check this one? My answers are kind of weird.

Answer 6

ouldn't it just be easier to put it into exponential form...

Answer 7

It is not loge

Answer 8

\[2 = \log(x^2 - 7x + 12) \implies 10^2 = x^2 - 7x + 12 \implies 0 = x^2 - 7x - 88\]\[x = \frac{-b ± \sqrt{b^2 - 4ac}}{2a}\]a = 1
b = -7
c = -88

\[x = \frac{7 ± \sqrt{49 + 352}}{2}\]Can you solve from there?

Answer 9

So you prefer base 10 :P

Answer 10

Well, it's written in base 10...

Answer 11

Where?

Answer 12

When there is no base denoted, it's assumed to be base 10. It's called a common logarithm.

Answer 13

I see, I thought it would be \(e\).. thanks for the info!

Answer 14

Thank you so much!!!

Answer 15

@zepp No problem :) I probably will get more used to e as I get farther into calculus.
@hooks4 You can just use a calculator for the rest. Split it into two expressions.
\[x = \frac{7 + \sqrt{401}}{2}\space and \space x = \frac{7 - \sqrt{401}}{2}\]

Answer 16

Wait, you got it from here?

Answer 17

Thank you @Calcmathlete and @zepp

Answer 18

np :)