Question

solve: log(x-1)+log(x-3)=8

Answer 1

Use:\[\log(a) +\log(b) = \log(ab)\]

Answer 2

\[a = (x-1) ~,~ b=(x-3)\]

Answer 3

and what do i do with the 8?

Answer 4

Just worry about the left side for now. What do you gt on the left side?

Answer 5

Good.

Answer 6

\[\log(x-1) +\log(x-3)= \log(x^2-4x+3)\]

Now you have \[\log(x^2-4x+3) =8\]To get rid of the log function and solve for the variables, we can raise both sides of this function by 10.\[\large 10^{\log(x^2-4x+3)}=10^8\]

Answer 7

We are then left with : \(x^2-4x+3 = 10^8\)

Answer 8

Oh I made a big mistake.

Answer 9

Do you see what I did wrong?

Answer 10

Remember: \(\large e^{\log_e(x)} = x~ , ~\log_e(e^x) =x\)

Answer 11

So instead of raising both sides by 10, I should have actually raised both sides by base \(e\) to get rid of the log function.

Answer 12

\[\large \log(x-1) +\log(x-3)= \log(x^2-4x+3)\]\[\large \log(x^2-4x+3) = 8\]\[\large e^{\log(x^2-4x+3)} = e^8\]\[\large x^2-4x+3=e^8\]

Answer 13

Can you solve it from here, or would you like me to keep going?

Answer 14

I would appreciate it, if you kept going. This concept is still confusing/new to me. Thank you so much for your help so far

Answer 15

Are you confused about any step?

Answer 16

In particular, I mean.

Answer 17

Not right now. Another side question: if it was written with "ln" instead of "log", would I still use "e"?

Answer 18

can you take a screenshot of actual question and post it ?

Answer 19

in school text books, when base is omitted log generally refers to base 10
and ln always refers to base e

Answer 20

Ah, ok. I did not know how to explain it.

Answer 21

for school textbooks :
\[\ln (x) = \log_e (x)\]

\[\log(x) = \log_{10}(x)\]

Answer 22

i meant. for example if the equation was: ln(x-1+ln(x-3)=8. Will I still use base e in the equation?