# Exponential and Logarithmic Equations

• Jun 15th 2006, 05:55 AM
kbryant05
Exponential and Logarithmic Equations
Good Morning,

My problem is: Use a calculator with a y^x key or ^ key to solve the following.

The 1986 explosion at Chernobyl nuclear power plant in the former Soviet Union sent about 1000 kilograms of radioactive cesium-137 into the atmosphere. The functon f(x)=1000(0.5)^ x over 30 describes the amount, f(x), in kilograms of cesium-137 remaining in Chernobyl x years after 1986. If even 100 kilograms of cesium-137 remain in Chernobyls atmosphere, the area is considered unsafe for human habitation. Find f(80) and determine if Chernobyl will be safe for human habitation by 2066.
• Jun 15th 2006, 06:28 AM
earboth
Quote:

Originally Posted by kbryant05
Good Morning,

My problem is: Use a calculator with a y^x key or ^ key to solve the following.

The 1986 explosion at Chernobyl nuclear power plant in the former Soviet Union sent about 1000 kilograms of radioactive cesium-137 into the atmosphere. The functon f(x)=1000(0.5)^ x over 30 describes the amount, f(x), in kilograms of cesium-137 remaining in Chernobyl x years after 1986. If even 100 kilograms of cesium-137 remain in Chernobyls atmosphere, the area is considered unsafe for human habitation. Find f(80) and determine if Chernobyl will be safe for human habitation by 2066.

Hello,

I hope that you are allowed to use the ln-button too.

Plug in the value(s) you know into the equation:

$\displaystyle f(x)=1000 \cdot \left( \frac{1}{2} \right)^{\frac{x}{30}$

That means you have to solve the equation for x:

$\displaystyle 80=1000 \cdot \left( \frac{1}{2} \right)^{\frac{x}{30}$

The next steps should be:

$\displaystyle 0.08=\left( \frac{1}{2} \right)^{\frac{x}{30}$

The exponent x/30 is a logarithm to the base 0.5. Use the base-change-formula (I am not sure, if this is the right expression. But maybe you recognize the name when you have a look at the solution)

$\displaystyle \frac{\ln(0.08)}{\ln(0.5)}=\frac{x}{30}$

So:

$\displaystyle 30\cdot\frac{\ln(0.08)}{\ln(0.5)}=x$. That means x is round about 110 years. In 2066 Chernobyl will be one of the most dangerous places on earth.

Greetings

EB
• Jun 15th 2006, 06:42 AM
kbryant05
Thanks Earboth. I don't think that answer is very comforting for anyone living in that area. It's actually really frightening.
Yes we are allowed to use the in button too!! ;)
Have a great day!
• Jun 15th 2006, 12:08 PM
Quick
Quote:

Originally Posted by earboth
That means x is round about 110 years. In 2066 Chernobyl will be one of the most dangerous places on earth.

Greetings

EB

I don't agree. You said that $\displaystyle f(x)=f(80)=80$, which would solve the equation for "how many years does it take to reach 80kg" but the question says it is $\displaystyle f(x)=f(80)$ so $\displaystyle x=80$

So you would substitute 80 for x instead of the function of x.

$\displaystyle f(x)=1000(0.5)^{\frac{x}{30}}$

$\displaystyle f(80)=1000(0.5)^{\frac{80}{30}}$

$\displaystyle f(80)=1000(0.5)^{\frac{8}{3}}$

$\displaystyle f(80)=1000(0.157490131236859)$

$\displaystyle f(80)=157.490131236859$

So, yes it is still lethal.