1. ## Logarthim/Exponential help..

Very quick one this:

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

I am sure the above is correct, but does:

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

(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 $\displaystyle (x+7)^3$ but have been given $\displaystyle x+7$ 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!

2. $\displaystyle e^{\ln{f(x)}} = f(x)$ and $\displaystyle e^{g(x)\cdot \ln{f(x)}} = e^{\ln{f(x)}^{g(x)}} = f(x)^{g(x)}.$
In your case we have $\displaystyle e^{3\ln(x+7)} = e^{\ln(x+7)^3} = (x+7)^3$.
So you're correct; the online calculator is wrong.

3. O.K - I was hoping it would have been x+7 though,

Now I've got a really difficult equation to solve...

4. 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.

5. No - I am pleased with being correct.

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

Somethign like:

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

Because it doesn't look simpler to me!

6. Then go the other way: 2ln(x+ 2)- 2ln(x+ 9)+ 2ln(x-1)+ 2ln(x+ 1). Does that look simpler to you?

7. Halls of Ivy, that's what I started with originally:

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

but simplified it to:

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

The post before this was slightly incorrect (exponent of 2 not 3).

8. Consider the two possibilities, let this expression be equal to y

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

which of those looks easier to simplify?

9. The actual full question is:

solve for y and simplfy as far as possible.

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

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

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

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

which of those looks easier to simplify?
The first: $\displaystyle e^y = e^{\ln[(x+2)^2(x+9)^-^2(x^2-1)^3]}=(x+2)^2(x+9)^-^2(x^2-1)^3$ ?

11. Which is the same as:

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

?

12. Originally Posted by MaverickUK82
The actual full question is:

solve for y and simplfy as far as possible.

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

I can separate y on the left by changing $\displaystyle 3ln(y+4)$ to $\displaystyle ln(y+4)^3$, 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

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

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

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

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

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

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

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

You can use the laws of exponents to simplify the RHS

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

13. Originally Posted by MaverickUK82
Which is the same as:

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

?
Yes

14. 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.

15. 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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