# Thread: logarithm

1. well you are almost to the solution:

convert the base to 10
log10 (x^2 + 4x + 3) / log10 3 = 1

log10 (x^2 + 4x + 3) = log10 3

comparing left and right hand side:
x^2 + 4x + 3 = 3

solving for x you will get x = 0, x = -4

then you substitute for each value of x into the main equation and carry out your reasoning why which value of x is acceptable and which is not acceptable

2. okay.. now i have 3 more problems on the studyguide that i just don't understand.. i really am trying though..

1... log x-3log 2=2... i know the answer is 800, and i have no idae where to start.. and the x and the 2 are regular, they are not subscripts.

2... log(6x+4)=2... the farthest i got was to put log in front of the 2..

3... ln(x-2) + ln(x-5) = ln(x-1).. i don't know where to start there, but i know it involves some e functions, i think.!  3. For the first one, well transform -3log2 into $\displaystyle -\log 2^3$
Then, as it is log, it's base 10. So $\displaystyle 2=2*1=2*\log(10)=\log(10^2)$

Hence $\displaystyle \log(x)-3\log(2)=2 \longleftrightarrow \log(x)-\log(8)=\log(100)$

Can you continue ? I've given you all the formulas you needed $\displaystyle log(6x+4)=2$

$\displaystyle log(6x+4)=log(10^2) \longleftrightarrow 6x+4=\dots$

ln(x-2) + ln(x-5) = ln(x-1)

group the logarithms in the left hand side, then solve.
If you want to go with exp, then :

$\displaystyle exp(ln(x-2)+ln(x-5))=exp(ln(x-1))$

$\displaystyle =exp(ln(x-2))*exp(ln(x-5))=exp(ln(x-1))$

$\displaystyle (x-2)(x-5)=(x-1)$

4. what is exp?

5. e power The e you talked about, exponential

6. i'm so lost on this last one... it's not even funny! i don't understand it, at all!

7. Ok so do it the right way :

ln(x-2)+ln(x-5)=ln[(x-2)(x-5)]

8. okay so what do you do with the ln(x+1) thats on the right side of the equal sign

9. If $\displaystyle ln(a)=ln(b)$ then a=b

According to it, you can say that (x-2)(x-5)=... ?

10. okay so after you work it out,

you get x^2-4x-11... right?

and then because you can't factor that, use the quadratic equation???

11. Hm...

I get (x-2)(x-5)=x-1
This makes x²-7x+10=x-1

x²-8x+11=0

This can be factored, yep, use the quadratic thing :

x²-8x+16-5=0

(x-4)²=5

12. okay well it says the answer is { 6.236 }
(1.764 is NOT a solution
we cannot take the log
of a negative number)

i don't understand that...

13. Since you have ln(x-2), ln(x-5) and ln(x-1), you must have :

x-2>0 <=> x>2
x-5>0 <=> x>5
x-1>0 <=> x>1

So finally, x has to be superior to 5. Hence 6.236 works, but not 1.764 14. but how can you get the 6 number??

15. (x-4)²=5

<=> $\displaystyle x-4=\sqrt{5}$ or $\displaystyle x-4=-\sqrt{5}$

<=> $\displaystyle x=4+\sqrt{5}$ or $\displaystyle x=4-\sqrt{5}$

And here is the miracle : $\displaystyle \sqrt{5} \approx 2.23606797749979$

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