Exponential and Logarithmic Functions (1 Viewer)

[FDownes]

Damn, I hate it when I get stuck on a question that should be incredibly simple. Can someone quickly help me out of this? Here's the question;

If log₇2 = 0.36 and log₇3 = 0.56, find the value of log₇14 and the value of log₇3.5.

 

[FDownes]

9 + 4 = 14?

EDIT: Beat'd

 

[tacogym27101990]

hahaha i can be about the biggest idiot.
guess u can discount everything i just sed
minds on english
sorry

 

[FDownes]

No worrys, thanks anyway.

EDIT: Oh, and BTW, I'm stuck on the next problem too. I'll add it to the opening post.

 

[midifile]

If log₇2 = 0.36 and log₇3 = 0.56, find the value of log₇14.

14 is 9 +4
so u use your log laws
log₇14 = log₇9 + log₇4
= log₇3² + log₇2²
=2log₇3 + 2log₇2
=2*0.36 + 2*0.56
=1.84
.... I hope

Your first log law is wrong
When you add two logs you times their numbers, so log₇9 + log₇4 = log₇(4x9) = log₇36.... plus either way, 14 = 9 + 5 not 9 + 4

However what you can do is say:
log₇14 = log₇(2x7)
= log₇2 + log₇7
= 0.36 + 1 (because log_aa = 1)
= 1.36
... I think thats right. I dont know what you needed the log₇3 for

 

[tacogym27101990]

yes yes i did totally screw that fella up.
owell

 

[FDownes]

Oooooooooooooooh... I can't believe I didn't think of that!

Thanks a heap!

 

[midifile]

Oooooooooooooooh... I can't believe I didn't think of that!

Thanks a heap!

no worries.

The other one is log₇(7/2) = log₇7 - log₇2
= 1 - 0.36
= 0.64

 

[Mark576]

Q. If log₇2 = 0.36 and log₇3 = 0.56, find the value of log₇14.

log₇14 = log₇(2*7) = log₇2+log₇7 = 1+0.36 = 1.36

 

[FDownes]

Ugh... I'm having a bad time with these questions. Here's another one;

Find the area enclosed between the curve y = ln x, the y-axis and the lines y = 1 and y = 3.

 

[webby234]

x = e^y (Make the subject x because you're finding area between y axis and curve)

So integrate that from 1 to 3.

Int(e^ydy)
= e^y
= e³ - e

 

[FDownes]

Okay, new question;

Find the exact gradient of the tangent to the curve y = e^{x + log_ex} at the point where x = 1.

This is my working;
= y' = 1/x * e^{x + log_ex}
= m = 1/(1) * e^{(1) + log_e(1)}
= m = 1 * e^{1 + 1}
= m = e²

Only according to the book, that's wrong. According to it, the answer should be 2e. Where have I gone wrong?

 

[Mark576]

y = e^{x + log_ex}
y' = (1+1/x)e^{x + log_ex}
At x = 1;
y' = 2e^{1 + log_e1} = 2e, since log_e1 = 0.

 

[FDownes]

Whoops, silly mistake. I was thinking log_e1 = 1.

Thanks.

 

[FDownes]

Another one;

Use simpson's rule with 5 function values to find the volume of the solid formed when the curve y = e^x is rotated about the y-axis from y = 3 to y = 5, correct to 2 significant figures.

My problem isn't so much in applying simpsons rule (that's really just substituting values in to a formula, not really very hard) but in figuring out how to get x² from e^x. I'm sure its fairly simple, but could someone explain?

 

[cwag]

y = e^x

from the definition of a log. we find log_a b = c changes to a^c = b

therefore.. log_e y = x
therefore x² = (ln y)²

 

[FDownes]

Ah... Right. Silly mistake again.

EDIT: Hmm... How would I go about integrating (ln y)²? Would it be 1/2(ln y) or am I completely off the track?

EDIT x 2: Oh crap, I'm completely forgetting my log laws again. (ln y)² = 2ln y, right? So how do I go about integrating that?

 

[foram]

Ah... Right. Silly mistake again.

EDIT: Hmm... How would I go about integrating (ln y)²? Would it be 1/2(ln y) or am I completely off the track?

EDIT x 2: Oh crap, I'm completely forgetting my log laws again. (ln y)² = 2ln y, right? So how do I go about integrating that?

ln y^2 = 2.ln y

but is that also true for (ln y)^2 ?

 

[FDownes]

Another question;

Find ∫x²e^{3 - 1} dx.

Don't forget the question in my last post though, I still need help with that.

EDIT: ^ I'm really not sure. I don't suppose you could go through it...?

 

[foram]

i don't think you can intergrate log, unless you use the simpsons rule or somthing like that, but thats not really intergrating.