# Logarthim/Exponential help..

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• Jan 19th 2011, 05:15 AM
MaverickUK82
Logarthim/Exponential help..
Very quick one this:

$
3ln(x+7) = ln(x+7)^3
$

I am sure the above is correct, but does:

$
exp^ (ln(x+7)^3) = (x+7)^3 or (x+7)
$

(Please excuse the wrong sized bracket after the exp function - I can't seem to get it to go the right size in LaTeX)

I am assuming it's $(x+7)^3$ but have been given $x+7$ as an answer from an online calculator.

Please help - I'm rubbish at logarithms and need to improve asap due to a home study course coming up!
• Jan 19th 2011, 05:22 AM
TheCoffeeMachine
$e^{\ln{f(x)}} = f(x)$ and $e^{g(x)\cdot \ln{f(x)}} = e^{\ln{f(x)}^{g(x)}} = f(x)^{g(x)}.$
In your case we have $e^{3\ln(x+7)} = e^{\ln(x+7)^3} = (x+7)^3$.
So you're correct; the online calculator is wrong.
• Jan 19th 2011, 05:35 AM
MaverickUK82
O.K - I was hoping it would have been x+7 though,

Now I've got a really difficult equation to solve...
• Jan 19th 2011, 05:38 AM
TheCoffeeMachine
Quote:

Originally Posted by MaverickUK82
O.K - I was hoping it would have been x+7 though,

Good luck with solving the big equation you've got.
• Jan 19th 2011, 06:11 AM
MaverickUK82
No - I am pleased with being correct.

Its just now i have a long equation to simplify - how would you go about the following:
$
ln(x+2)^2 + ln(x+9)^-^2 + ln(x^2-1)^2
$

Somethign like:

$
ln((x+2)^2(x+9)^-^2(x^2-1)^2)
$

Because it doesn't look simpler to me!
• Jan 19th 2011, 06:15 AM
HallsofIvy
Then go the other way: 2ln(x+ 2)- 2ln(x+ 9)+ 2ln(x-1)+ 2ln(x+ 1). Does that look simpler to you?
• Jan 19th 2011, 06:36 AM
MaverickUK82
Halls of Ivy, that's what I started with originally:

$2ln(x+2)-2ln(x+9)+3ln(x^2-1)$

but simplified it to:

$ln(x+2)^2 + ln(x+9)^-^2+ln(x^2-1)^3$

The post before this was slightly incorrect (exponent of 2 not 3).
• Jan 19th 2011, 06:41 AM
e^(i*pi)
Consider the two possibilities, let this expression be equal to y

$e^y = e^{\ln[(x+2)^2(x+9)^-^2(x^2-1)^3]}$ OR $e^y = e^{ln(x+2)^2 + ln(x+9)^-^2+ln(x^2-1)^3}$

which of those looks easier to simplify?
• Jan 19th 2011, 06:42 AM
MaverickUK82
The actual full question is:

solve for y and simplfy as far as possible.

$3ln(y+4)=ln(x+2)^2 + ln(x+9)^-^2+ln(x^2-1)^3$

I can separate y on the left by changing $3ln(y+4)$ to $ln(y+4)^3$, using the exponential rule and taking cube root and moving over the 4, but it's the simplification I'm struggling with.
• Jan 19th 2011, 06:46 AM
MaverickUK82
Quote:

Originally Posted by e^(i*pi)
Consider the two possibilities, let this expression be equal to y

$e^y = e^{\ln[(x+2)^2(x+9)^-^2(x^2-1)^3]}$ OR $e^y = e^{ln(x+2)^2 + ln(x+9)^-^2+ln(x^2-1)^3}$

which of those looks easier to simplify?

The first: $e^y = e^{\ln[(x+2)^2(x+9)^-^2(x^2-1)^3]}=(x+2)^2(x+9)^-^2(x^2-1)^3$ ?
• Jan 19th 2011, 06:52 AM
MaverickUK82
Which is the same as:

$\frac{(x+2)^2(x^2-1)^3}{(x+9)^2}$

?
• Jan 19th 2011, 06:52 AM
e^(i*pi)
Quote:

Originally Posted by MaverickUK82
The actual full question is:

solve for y and simplfy as far as possible.

$3ln(y+4)=ln(x+2)^2 + ln(x+9)^-^2+ln(x^2-1)^3$

I can separate y on the left by changing $3ln(y+4)$ to $ln(y+4)^3$, using the exponential rule and taking cube root and moving over the 4, but it's the simplification I'm struggling with.

You'd be better off clearing the log terms first

Use what you did in post 4

$\ln(y+4)^3 = \ln[(x+2)^2(x+9)^-^2(x^2-1)^2]$

Those logarithms will cancel to give $(y+4)^3 = \dfrac{(x+2)^2(x^2-1)^2}{(x+9)^3}$

You can take the cube root now and remember that $\sqrt[3]{ab} = \sqrt[3]{a} \cdot \sqrt[3]{b}$

=================================

If you want to use the form in your last post you will get

$ln(y+4)^3=ln(x+2)^2 + ln(x+9)^-^2+ln(x^2-1)^3$

$e^{ln(y+4)^3}=e^{ln(x+2)^2 + ln(x+9)^-^2+ln(x^2-1)^3}$

You can use the laws of exponents to simplify the RHS

$(y+4)^3 = e^{\ln(x+2)^2} \cdot e^{\ln(x+9)^{-2}} + e^{\ln(x^2-1)^3} = (x+2)^2 \cdot (x+9)^{-2} \cdot (x^2-1)^3$
• Jan 19th 2011, 06:52 AM
e^(i*pi)
Quote:

Originally Posted by MaverickUK82
Which is the same as:

$\frac{(x+2)^2(x^2-1)^3}{(x+9)^2}$

?

Yes
• Jan 19th 2011, 07:08 AM
MaverickUK82
Ah - fantastic - so I was on the right tracks - tricky little things - logarithms.

What did you mean whne you said "Those logarithms will cancel to give" I have not heard that term given with regards to logarithms.
• Jan 19th 2011, 07:32 AM
e^(i*pi)
Quote:

Originally Posted by MaverickUK82
Ah - fantastic - so I was on the right tracks - tricky little things - logarithms.

What did you mean whne you said "Those logarithms will cancel to give" I have not heard that term given with regards to logarithms.

I just meant that because the base of the logarithm is the same (base e) so too must the expressions inside them be the same. Cancels is probably pretty dodgy terminology for logs lol
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