# Thread: Quick Algebra question (logs and stuff)

1. i get a negative number every time i try to solve...

2. Let's do it this way then.

Rewrite $e^{\frac{ln(3/5)t}{5}}=(3/5)^{\frac{t}{5}}$

$50=100(3/5)^{\frac{t}{5}}$

$1/2=(3/5)^{\frac{t}{5}}$

$ln(1/2)=\frac{t}{5}ln(3/5)$

$t=\frac{5ln(1/2)}{ln(3/5)}=6.78$

3. ooooooh

4. Originally Posted by reynardin
wait, so is K supposed to equal the $

\frac{\ln(\frac{3}{5})}{5}=\frac{\ln\bigg(\bigg(\f rac{5}{3}\bigg)^{-1}\bigg)}{5}=\frac{-1\ln(\frac{5}{3})}{5}
$
?
Yes. That is the decay constant you need.

5. ok, i suppose I get it now.

Next problem.

$\log_3(x)-\log_4(7)=\log_7(5)$

I put the equation into logs of base 10, but I don't know what to do after that.

6. Hello,

Originally Posted by reynardin
ok, i suppose I get it now.

Next problem.

$\log_3(x)-\log_4(7)=\log_7(5)$

I put the equation into logs of base 10, but I don't know what to do after that.
Of base 10 ? o.O

Put it in the exponential logarithm... $\log_x (y)=\frac{\ln(y)}{\ln(x)}$

---> $\frac{\ln x}{\ln 3}-\frac{\ln 7}{\ln 4}=\frac{\ln 5}{\ln 7}$

--> $\ln x-\frac{\ln 7 \cdot \ln 3}{\ln 4}=\frac{\ln 5 \cdot \ln 3}{\ln 7}$

$\ln x=\frac{\ln 5 \cdot \ln 3}{\ln 7}+\frac{\ln 7 \cdot \ln 3}{\ln 4}$

---> $x=exp\left(\frac{\ln 5 \cdot \ln 3}{\ln 7}+\frac{\ln 7 \cdot \ln 3}{\ln 4}\right) \approx 11.59699220980753$

7. EH, thats what i mean >.<
lol
as in, log base ten of x

8. BLAH ignore last post

9. what does the exp mean?

10. Originally Posted by reynardin
what does the exp mean?
$exp(\dots)$ is like $e^{\dots}$

11. ah ok

12. anyways,
I LOVE YOU GUYS!

done for now =D

13. Originally Posted by galactus
Let's do it this way then.

Rewrite $e^{\frac{ln(3/5)t}{5}}=(3/5)^{\frac{t}{5}}$

$50=100(3/5)^{\frac{t}{5}}$

$1/2=(3/5)^{\frac{t}{5}}$

$ln(1/2)=\frac{t}{5}ln(3/5)$

$t=\frac{5ln(1/2)}{ln(3/5)}=6.78$
btw, how does he get $e^{\frac{ln(3/5)t}{5}}=(3/5)^{\frac{t}{5}}$? I get that ln and e basically "cancel" each other out, but why is the 3/5 to the power of t/5?

14. Disregard my last post, i figured it out.
Originally Posted by Moo
Hello,
...
Put it in the exponential logarithm... $\log_x (y)=\frac{\ln(y)}{\ln(x)}$

---> $\frac{\ln x}{\ln 3}-\frac{\ln 7}{\ln 4}=\frac{\ln 5}{\ln 7}$

--> $\ln x-\frac{\ln 7 \cdot \ln 3}{\ln 4}=\frac{\ln 5 \cdot \ln 3}{\ln 7}$

$\ln x=\frac{\ln 5 \cdot \ln 3}{\ln 7}+\frac{\ln 7 \cdot \ln 3}{\ln 4}$

---> $x=exp\left(\frac{\ln 5 \cdot \ln 3}{\ln 7}+\frac{\ln 7 \cdot \ln 3}{\ln 4}\right) \approx 11.59699220980753$
does multiplying an equation such as $\ln(x)=....$ by e make it into $x=e*...$?

15. Originally Posted by reynardin
Disregard my last post, i figured it out.

does multiplying an equation such as $\ln(x)=....$ by e make it into $x=e*...$?
For the last one

$a^{bc}=(a^b)^c$

So $e^{\ln(\frac{3}{5})\frac{t}{5}}=(e^{\ln(\frac{3}{5 })})^{\frac{t}{5}}=\bigg(\frac{3}{5}\bigg)^{\frac{ t}{5}}$

and no multiplying by it does nothign

raising e to each side power does

for example $\ln(45x-220)=\ln(5+17x)$

If you multiplied both sides by e would do nothing

but raising e to eachside would give

$e^{\ln*45x-220)}=e^{\ln(5+17x)}\Rightarrow{45x-220=5+17}$

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