Question

PLEASE HELP : I need help of intergral exponents THANKS IN ADVANCE FOR ANY HELP ;) (i know A but not B)

Answer 1

all you need to know about logarithms
https://www.khanacademy.org/math/algebra/logarithms-tutorial/logarithm_properties/v/introduction-to-logarithm-properties

Answer 2

No logs are needed here, only optimization.
To find the maximum height, find d/dx(3(.07)^n) and set it equal to 0, then solve for x. After that, determine by either using a number line or by looking at a graph of h(x) what values of x are maximums or minimums.

Answer 3

Hmm logs is probably a better way to do it :3
But yah that's a clever idea also :)

Answer 4

you don't need a derivative. just set h to be .5. solve for n. and yes you need a log.

Answer 5

You would need logs if you were looking for the number n of bounces at which h(x)=.5, but if you do optimization you don't need logs XD
good point

Answer 6

yeah because i tryed to solve for n but since n as an exponent i had no idea how

Answer 7

\[\large h=3(0.7)^n\]We want to solve for \(n\) when \(h=\dfrac{1}{2}\).

\[\large \frac{1}{2}=3(0.7)^n\]We'll start by taking the natural log of both sides,\[\large \ln\left(\frac{1}{2}\right)=\ln\left[3(0.7)^n\right]\]

Answer 8

h = 3*(.07)^n
.5/3 = .07^n
ln(.5/3) = ln(.07^n)
ln(.5/3) = n*ln(.07)
[ln(.5/3)]/ln(.07) = n

Answer 9

Now we need to apply a familiar rule of logs,\[\large \color{royalblue}{\log(ab)=\log (a)+\log (b)}\]

We'll do this to split up the exponential term and the 3.\[\large \ln\left(\frac{1}{2}\right)=\ln3+\ln(0.7)^n\]

Answer 10

Understand that step? It's a little tricky if you don't remember your log rules :)

Answer 11

Oh my :U

Answer 12

Well just try to remember the blue formula I pasted above :)
When you have 2 things being multiplied inside of one log, you can rewrite it as the sum of two logs.

Answer 13

\[\large \ln\color{salmon}{3}\color{royalblue}{(0.7)^n} \qquad = \qquad \ln\color{salmon}{3}+\ln\color{royalblue}{(0.7)^n}\]

Answer 14

There's a few more tricky steps ^^ hehe

Answer 15

\[\large \ln\left(\frac{1}{2}\right)=\ln3+\ln(0.7)^n\]
We need to apply another rule of logs,\[\large \color{royalblue}{\log(a^b)=b \log (a)}\]This rule allows you to bring a power down outside of a log as a factor in front.
This is the whole reason we applied the log in the first place. Because it allows us to take the \(n\) out of the exponent position.

Answer 16

\[\large \ln (0.7)^{\color{salmon}{n}} \qquad \rightarrow \qquad \color{salmon}{n}\ln(0.7)\]