# Math Help - Logarithms (advanced)

1. ## Logarithms (advanced)

When it comes to math i usually have an indication as to what I am supposed to do but this task was completely impossible for me.

M(t)=53e-t/223

t=amount of years. I am supposed to calculate when the function has decreased by 1/3 of it's original size using logarithmical laws. This time I can't show you my steps because I don't even know where to begin!

I also have this problem that i find equally difficult.

(ex-5)/(5e-x-4)=2

I don't even know where to begin and it's driving me absolutely insane lol, would a kind soul please help me ?

2. ## Re: Logarithms (advanced)

$M(t)=53e^{-t/223}$

let the original size be at $t=0$

$53e^{0} = 53=M_0$

we want to find $t$ such that

$M(t)=\dfrac {M_0}{3} = M_0 e^{-t/223}$

$\dfrac 1 3 = e^{-t/223}$

$\ln(1/3) = -t/223$

$t = -223 \ln(1/3) \approx 245$

For the second problem

$\dfrac {e^x-5}{5e^{-x}-4}=2$

$e^x-5 = 10e^{-x}-8$

$e^x =10e^{-x}-3$

Now a trick. Multiply everything by $e^x$ and get all the terms on one side of the equation.

$e^{2x} + 3e^x -10=0$

Now let $u=e^x$ and we transform this to a quadratic equation in $u$.

$u^2 + 3u - 10=0$

$(u+5)(u-2)=0$

$u=5 \vee u=-2$

$x=\ln(u)$

$\ln(-5)$ is not defined for real numbers so we ignore -5. Thus

$x=\ln(2)$

3. ## Re: Logarithms (advanced)

Thank you so, so much for the help! but i tried the i tried the calculation of the first one and it's apperantly incorrect it says that the awnser is supposed to be x=(a)ln(b) where a and b are whole numbers and where a can't asume the value of 1

4. ## Re: Logarithms (advanced)

Originally Posted by Foxtrot
Thank you so, so much for the help! but i tried the i tried the calculation of the first one and it's apperantly incorrect it says that the awnser is supposed to be x=(a)ln(b) where a and b are whole numbers and where a can't asume the value of 1
You're correct. I screwed up.

$M(t)=53e^{-t/223}$

let the original size be at $t=0$

$53e^{0} = 53=M_0$

we want to find $t$ such that

$M(t)=\dfrac {M_0}{3} = M_0 e^{-t/223}$

$\dfrac 1 3 = e^{-t/223}$

Apply $\ln(1/x)=-\ln(x)$

$\ln(3) = t/223$

$t = 223 \ln(3)$

The correct answer is $223\ln(3)$

5. ## Re: Logarithms (advanced)

Originally Posted by romsek
Originally Posted by Foxtrot
Thank you so, so much for the help! but i tried the i tried the calculation of the first one and it's apperantly incorrect it says that the awnser is supposed to be x=(a)ln(b) where a and b are whole numbers and where a can't asume the value of 1
You're correct. I screwed up.
Maybe it's the beer talking, but as far as I'm concerned, there was no screw up on the psart of Sir romsek. If there was a screw up, I'd say it was on Foxtrot's part for not including that condition on his/her first post in the first place.