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elaine · « **on:** April 26, 2008, 07:27:17 pm »

darn loggies are killing me again.

could someone help with these questions?

thanks

Solve for x

1. $\log_{e} y$ = $(\log_{e} x)$ $+ 2$

2. ln(y) = 3- ln(x)

3. $2^{x}$ = $3^{(x+1)}$

Find the exact value for x, expressing your answer in terms of logarithms base e

4. $4e^{3x}= 287$ (can i do this without a calculator?)

5. Find the range where y= f[g(x)] exists and state the maximal domain where

f(x)= ln (2-x)
g(x)= |x|

6. The expression $3^\log_{3} (x-4)$ equals?

Also, just to clarify- If i wanted to get rid of the $\log_{10}$ in $\log_{10} 3x$ , can I raise the expression by 10.

i.e. $10^\log_{10} 3x}$ to get $3x$ or is that illegal

Mao · « **Reply #1 on:** April 26, 2008, 08:00:32 pm »

1.
$log_e(y)=log_e(x)+2$

$log_e(y)=log_e(x)+log_e\left(e^2\right)$

$log_e(y)=log_e\left(e^2\cdot x\right)$

$y=e^2\cdot x$

2. similar principle =)

3.
$2^x=3^{x+1}$

$\left(e^{log_e 2}\right)^x = \left(e^{log_e 3}\right)^{x+1}$

$e^{log_e(2)\cdot x}=e^{log_e(3)\cdot (x+1)}$

4. grrrr *killself*

5.
the implied domain for f is that $2-x>0$ , hence $x<2$

since this is a compound function, f takes the range of g, that is:

$|x|<2$
which is evaluated to
$-2<x<2$
this is the maximal domain.

to find the range, we go back one step to $|x|<2$ , that is, the minimum input into f is 0, and the maximum is 2 (not including)

since f is well behaved and does not change direction (logarithmic functions does not change directions), we can simply substitude in 0 and 2 to find its range:

$f(0)=ln(2)$ , and as x approaches 2, we get $\lim_{x \to 2}ln(2-x) = \lim_{u \to 0}ln(0) = -\infty$

so our range is $(-\infty,ln(2))$

6.
the question below it is well asked, and it is true.

$3^{log_3(x-4)} = x-4$

to be more generic:

if u is any expression:

$a^{log_a u} = log_a (a^u) = u$

it should be noted that logarithms and exponentials are inverse of each other, and by principle: $f(f^{-1}(x))=f^{-1}(f(x))=x$

=)

elaine · « **Reply #2 on:** April 26, 2008, 08:26:58 pm »

thanks mao, frikkin genius

/0 · « **Reply #3 on:** April 26, 2008, 09:46:32 pm »

Isn't 4 this?

$e^{3x}=\frac{287}{4}$

$3x=\log_e {\left(\frac{287}{4}\right)}$

$x=\frac{1}{3}\log_e {\left(\frac{287}{4}\right)}$

dcc · « **Reply #4 on:** April 26, 2008, 09:52:03 pm »

since the question states:

Find the exact value for x, expressing your answer in terms of logarithms base e

then yes, 4 is that

Mao · « **Reply #5 on:** April 26, 2008, 10:20:09 pm »

Quote from: dcc on April 26, 2008, 09:52:03 pm

since the question states:

Find the exact value for x, expressing your answer in terms of logarithms base e

then yes, 4 is that

oops, my bad

elaine · « **Reply #6 on:** April 26, 2008, 10:36:03 pm »

cool thanks divide

elaine · « **Reply #7 on:** April 27, 2008, 11:42:17 am »

hey mao, i don't get what to do with the rest of Q3, yeahh i suck, but what do i do next?

also how does $\ log_{e} 4e^{3x}$ become $\ log_{e} 4 + 3x$

Mao · « **Reply #8 on:** April 27, 2008, 11:49:55 am »

question 3:
$e^{log_e(2)\cdot x}=e^{log_e(3)\cdot (x+1)}$

$log_e(2)\cdot x = log_e(3)\cdot (x+1)$

and

$log_e4e^{3x}$

$\implies log_e 4 + log_e e^{3x}$

$\implies log_e 4 + 3x$

elaine · « **Reply #9 on:** April 27, 2008, 12:04:08 pm »

sorry, i'm a bit confused mao

the answer for question 3 is $x= \frac{\log_{e} 3}{\log_{e} (2/3)}$

sorry if i've confused you!

elaine · « **Reply #10 on:** April 27, 2008, 12:06:19 pm »

and btw- that's a new question:

$\ log_{e} 4e^{3x}$ become $\ log_{e} 4 + 3x$

i have no idea where you've got the $\ log_{e} 187$ from!!

elaine · « **Reply #11 on:** April 27, 2008, 12:11:24 pm »

huh? sorry, it's not question 4- it's a new one
i get question 4

Mao · « **Reply #12 on:** April 27, 2008, 12:12:11 pm »

question 3:
$2^x=3^{x+1}$

$\left(e^{log_e 2}\right)^x = \left(e^{log_e 3}\right)^{x+1}$

$e^{log_e(2)\cdot x}=e^{log_e(3)\cdot (x+1)}$

$log_e(2)\cdot x = log_e(3)\cdot (x+1)$

$log_e(2)\cdot x = log_e(3)\cdot x + log_e(3)$

$(log_e(2)-log_e(3)) \cdot x = log_e(3)$

$log_e\left(\frac{2}{3}\right) \cdot x = log_e(3)$

$x=\frac{log_e(3)}{log_e\left(\frac{2}{3}\right)}$

Quote from: elaine98 on April 27, 2008, 12:11:24 pm

huh? sorry, it's not question 4- it's a new one
i get question 4

ahhh. i'm going nuts

the numbers are exactly the same tho.... =\

elaine · « **Reply #13 on:** April 27, 2008, 12:19:54 pm »

ah cool, i get q3 now

thanks

Quote

ahhh. i'm going nuts

the numbers are exactly the same tho.... =\

but i promise- they're different questions! lol there's not even a 287!

maybe i'll rephrase the question

The expression $\log_{e} (4e^{3x})$ is equal to....

And the answer is $\log_{e} (4) + 3$

how do i get the answer?

Mao · « **Reply #14 on:** April 27, 2008, 12:24:06 pm »

Quote from: Mao on April 27, 2008, 11:49:55 am

$log_e4e^{3x}$

$\implies log_e 4 + log_e e^{3x}$

$\implies log_e 4 + 3x$

and u forgot an "x" up there in your answer =)

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