Exponential and Logarithmic Functions help!!!

• Oct 24th 2007, 04:52 PM
Exponential and Logarithmic Functions help!!!
Hi everyone!

I am currently having trouble understanding a couple parts of my homework. I filled in the rest but these got confusing:

C. Change each equation to its exponential form:

$lne^2$

D. Change each equation to its Log form. Assume y > 0 and b > 0.

$y = e^x$

Quote:

Originally Posted by ThePerfectHacker
If $y=e^x$ then $x = \ln y$.

E. Evaluate each log.

$lne^3$

Quote:

Originally Posted by ThePerfectHacker
$\ln e^n = n$ because $e^{(n)} = e^n$.

Now this is the part I am more confused about

F. Solve each exponential equation.

$0.3(4^0.2x) = 0.2$

$9^x - 3^(x+1) + 1 = 0$

$log2(3x + 2) - log4x = 3$

$(e^x + e^-x) / 2 = 3$

You can answer them for me but please explain to me how you got the answer or what I need to do to get the answer.

I'm still working on these problems and finished a few of them but I want to see if I am positively correct.

Thank you guys!

--Alex
• Oct 24th 2007, 05:54 PM
ThePerfectHacker
Quote:

Hi everyone!

I am currently having trouble understanding a couple parts of my homework. I filled in the rest but these got confusing:

C. Change each equation to its exponential form:

ln e^2

D. Change each equation to its Log form. Assume y > 0 and b > 0.

y = e^x

E. Evaluate each log.

ln e^3

Now this is the part I am more confused about

F. Solve each exponential equation.

0.3(4^0.2x) = 0.2

9^x - 3^(x+1) + 1 = 0

log2(3x + 2) - log4x = 3

(e^x + e^-x) / 2 = 3

You can answer them for me but please explain to me how you got the answer or what I need to do to get the answer.

I'm still working on these problems and finished a few of them but I want to see if I am positively correct.

Thank you guys!

--Alex

If $y=e^x$ then $x = \ln y$.

And $\ln e^n = n$ because $e^{(n)} = e^n$.
• Oct 24th 2007, 06:56 PM
Thank you PerfectHacker for the answers. :)

I looked at the answers and I figured how it worked. Thanks!

Anyone else know how to work with the other problems, especially section F?

edit: I edited the main post and added the answers and made the problems appear more clearer. :P
• Oct 24th 2007, 09:24 PM
Soroban
Hello, Alex!

Could you re-type part F?

Quote:

F. Solve each exponential equation.

$[1] \;\;0.3(4^0.2x) = 0.2$

Can't read it . . .

Is that: . $0.3(4^{0.2x}) \:=\:0.2$ ?

Quote:

$[2]\;\;9^x - 3^(x+1) + 1 = 0$
I assume that it is: . $9^x - 3^{x+1} + 1 \:=\:0$

You need to enclose exponents in brackets: .3^{x+1}

We have: . $\left(3^2\right)^x - 3^x\!\cdot\!3 + 1 \:=\:0$

. . then: . $\left(3^x\right)^2 - 3\left(3^x\right) + 1\:=\:0$

Let $u \,=\,3^x$ and we have: . $u^3 - 3u + 1 \:=\:0$

Quadratic Formula: . $u \;=\;\frac{3\pm\sqrt{3^2-4(1)(1)}}{2(1)} \;=\;\frac{3\pm\sqrt{5}}{2}$

Back-substitute: . $3^x \:=\:\frac{3\pm\sqrt{5}}{2}$

Therefore: . $x \;=\;\log_3\left(\frac{3\pm\sqrt{5}}{2}\right)$

Quote:

$[3]\;\;log2(3x + 2) - log4x = 3$
Use a back-slash (\) before "log" . . . It looks better.
Also you can make subscripts with _2 and _4

We have: . $\log_2(3x+2) - \log_4(x) \:=\:3$

Use the Base-change Formula on the second log:
. . $\log_4(x) \:=\:\frac{\log_2(x)}{\log_2(4)} \:=\:\frac{\log_2(x)}{2}$

The equation becomes: . $\log_2(3x+2) - \frac{1}{2}\log_2(x) \:=\:3$

Multiply by 2: . $2\log_2(3x+2) - \log_2(x) \:=\:6\quad\Rightarrow\quad \log_2(3x+2)^2 - \log_2(x) \:=\:6$

Then we have: . $\log_2\left[\frac{(3x+2)^2}{x}\right] \:=\:6$

Can you finish it now?

Quote:

$[4]\;\;(e^x + e^-x) / 2 = 3$

Multiply by 2: . $e^x + e^{-x} \:=\:6$

Multiply by $e^x\!:\;\;e^{2x} + 1 \:=\:6e^x\quad\Rightarrow\quad e^{2x} - 6e^x + 1\:=\:0$

And we have a quadratic: . $\left(e^x\right)^2 - 6\left(e^x\right) + 1 \:=\:0$

Proceed as we did in part [2].

• Oct 24th 2007, 09:35 PM
earboth
Quote:

F. Solve each exponential equation.
1. $0.3(4^0.2x) = 0.2$
2. $9^x - 3^(x+1) + 1 = 0$
3. $log2(3x + 2) - log4x = 3$
4. $(e^x + e^-x) / 2 = 3$
...

Hello,

to 1.:

$0.3(4^{0.2x}) = 0.2~\iff~4^{0.2x} = \frac23~\iff~x=5\cdot \log_{4}\left(\frac23\right)$

to 2.:

$9^x - 3^{x+1} + 1 = 0~\iff~3^{2x}-3\cdot 3^x+1=0$ Use $y=3^x$ to get $y^2-3y+1=0$. Solve for y. Don't forget to re-substitute to calculate x.

to 3.:

$\log_{2}(3x + 2) - \log_{4}(x) = 3~\iff~\log_{2}(3x+2)-\frac12 \cdot \log_{2}(x) = 3$ $~\iff~\log_{2}\left(\frac{3x+2}{\sqrt{x}}\right)=3$ Therefore $\frac{3x+2}{\sqrt{x}} = 8$ Solve for x.

to 4.:

$(e^x + e^{-x}) / 2 = 3~\iff~e^x+\frac1{e^x}=6~\iff~(e^x)^2+1=6e^x$ Use $y = e^x$ to get $y^2-6y+1=0$ . Solve for y. Don't forget to resubstitute to calculate x.