Log Question (1 Viewer)

[Mc_Meaney]

Hey guys, I am confused with a couple of log questions...they probably are really easy...but hey..

Question Log(7)2 = 0.36 and Log(7)5 = 0.83

ok so
(i) Log(7)20 and (ii) Log(7)50 are the questions I am having probs with, I have gotten all the others....just...rargh I dunno.
I have the answers in the back of my book, I just need some help on the working.
Thanx In Advance

NOTE numbers in brackets are meant to be little lol

 

[Riviet]

(i) log₇20 = log₇(2x2x5)
= log₇2 + log₇2 + log₇5
= 0.36 + 0.36 + 0.83
= 1.55

(ii) log₇50 = log₇(2x5x5)
= log₇2 + log₇5 + log₇5
= 0.36 + 0.83 + 0.83
= 2.02

 

[Mc_Meaney]

Thanx rivlet ur a champ

 

[Mc_Meaney]

2e(2t) = 75

e(3t) =5

How in gods name do you do these...

 

[Riviet]

(Note that ln means log_e)

2e^2t = 75
e^2t = 75/2
ln(e^2t) = ln(75/2)
2t.lne = ln(75/2)
2t = ln(75/2)
t = (1/2).ln(75/2)

e^3t = 5
ln(e^3t) = ln5
3t.lne = ln5
3t = ln5
t= (1/3).ln5

Use a calculator to evaluate these if necessary.

 

[Mc_Meaney]

Solve
e^(log3x) = 6

HELP!!!

 

[darkliight]

It's important to say what base your log is in (like your first question, your base was 7).

I assume you mean log to the base e (ln) though.

e^(ln x) = x

So e^(ln 3x) = 3x = 6, x = 2.

 

[Riviet]

Whenever you have e^log_eX, since the exponential and logarithm are inverse functions, they cancel each other out and you get X.

So e^log_e(3x)=3x

So 3x=6
x=2

 

[Mc_Meaney]

It's important to say what base your log is in (like your first question, your base was 7).

I assume you mean log to the base e (ln) though.

e^(ln x) = x

So e^(ln 3x) = 3x = 6, x = 2.

Thats how the question was written :burn: stupid thing

 

[Raginsheep]

If its not specified, assume its base e.

 

[Mc_Meaney]

Ive got this question right, I just need the reasoning confirmed:

2^log(2)64

So I assume you do it like this

Log(2)64
2(x) = 64
2(x) = 2(6)
x = 6

Then do
=2^6
=64

 

[Trev]

Yeah, well log₂64 = log₂2⁶ = 6log₂2 = 6.

 

[Mc_Meaney]

Cool. Thanks buddy