# Exponential and logarithmic

• Apr 8th 2008, 04:52 AM
chaneliman
Exponential and logarithmic
hey, i haven't been for quite a while

1. If you invest $100 at 2% per day, compounding daily, the amount of money you would have after x days is given by y=100(1.02)^x. For how many days would you have to invest to double your money. 2. How would you go about sketching 2* 10^(x/10)+4? 3. y=2^x, y=3^x and y= 5^x i. For what of value of x is 2^x > 3^x > 5^x ii. For what value of x is 2^x < 3^x < 5^x iii. For what value of x is 2^x = 3^x = 5^x Thanks • Apr 8th 2008, 05:10 AM earboth Quote: Originally Posted by chaneliman hey, i haven't been for quite a while 1. If you invest$100 at 2% per day, compounding daily, the amount of money you would have after x days is given by y=100(1.02)^x. For how many days would you have to invest to double your money.

2. How would you go about sketching 2* 10^(x/10)+4?

3. y=2^x, y=3^x and y= 5^x
i. For what of value of x is 2^x > 3^x > 5^x
ii. For what value of x is 2^x < 3^x < 5^x
iii. For what value of x is 2^x = 3^x = 5^x

Thanks

to #1:

Solve for x:

$200=100 \cdot (1.02)^x$ (I've got 35 days)

to #2:

$f(x)=2 \cdot 10^{\frac x{10}}+4 = 2 \left(\sqrt[10]{10} \right)^x+4$ with $\sqrt[10]{10} \approx 1.2589$

to # 3(iii):

$a^x = b^x~\wedge~b \neq a ~\implies~ x = 0$

(i) x < 0
(ii) x > 0
• Apr 8th 2008, 06:10 AM
janvdl
Quote:

Originally Posted by chaneliman
hey, i haven't been for quite a while

1. If you invest \$100 at 2% per day, compounding daily, the amount of money you would have after x days is given by y=100(1.02)^x. For how many days would you have to invest to double your money.

$100(1.02)^x = 200$

$(1.02)^x = 2$

$x = log_{1.02} 2$

$x$ is roughly 35, in other words, 35 days.

Quote:

Originally Posted by chaneliman
2. How would you go about sketching 2* 10^(x/10)+4?

You are supposed to already know the basic form of an exponential curve.

Let x = 0
Then you will find y = 6

So you know it intersects the y-axis at 6

We can make $2(10^{\frac{x}{10}})$ infinitely small, but not exactly zero.

So if that part is infinitely small, we have on constant (+ 4) which will be our asymptote. See the attached image for clarification.

Quote:

Originally Posted by chaneliman
3. y=2^x, y=3^x and y= 5^x
i. For what of value of x is 2^x > 3^x > 5^x
ii. For what value of x is 2^x < 3^x < 5^x
iii. For what value of x is 2^x = 3^x = 5^x

i) Where x < 0
ii) Where x > 0
iii) Where x = 0