Logarithmic and exponential function questions

• Mar 15th 2008, 05:55 AM
tyana21
Logarithmic and exponential function questions
log3(7x+6)-log94x(squared)=1

ey-1=6/e
ln x2 +ey=18, find the value of (square root)ey(ln x3)
• Mar 15th 2008, 06:26 AM
Peritus
$
\begin{gathered}
e^{y - 1} = \frac{6}
{e} \hfill \\
\Leftrightarrow e^{y - 1} = e^{\ln 6 - 1} \hfill \\
\Leftrightarrow y = \ln 6 \hfill \\
\end{gathered}
$

-----------------------------------------------------------
• Mar 15th 2008, 06:47 AM
Moo
Hello,

$\log_3 (7x+6) = \frac{\ln(7x+6)}{\ln(3)}$

$\log_9 (4x^2) = \frac{\ln(4x^2)}{\ln(9)}$

But we know that $\ln(a^b) = b \ln(a)$

So, as 4x² = (2x)², \ln(4x²)=2 \ln(2x) and as 9 = 3², \ln(9) = 2 \ln(3)

Hence

$\log_3 (7x+6) - \log_9 (4x^2) = \frac{\ln(7x+6)}{\ln(3)} - \frac{2 \ln(2x)}{2 \ln(3)}$ $= \frac{\ln(7x+6)}{\ln(3)} - \frac{\ln(2x)}{\ln(3)} = \frac{\ln(7x+6) - \ln(2x)}{\ln(3)} = 1$

But ln(a)-ln(b)=ln(a/b)

<=> $\ln(\frac{7x+6}{2x}) = \ln(3)$

You'll just have to study the values of x for which the equation can't be solved.
• Mar 15th 2008, 09:48 AM
Soroban
Hello, tyana21!

An I reading the first one correctly?

Quote:

$\log_3(7x+6) - \log_9(4x)^2 \:=\:1$

The first log can be written: . $\log_3(7x+6) \;=\;2\!\cdot\!\log_9(7x+6)$ .**

The second can be written: . $2\!\cdot\!\log_9(4x)$

The equation becomes: . $2\!\cdot\!\log_9(7x+6) - 2\!\cdot\!\log_9(4x) \;=\;1$

Divide by 2: . $\log_9(7x+6) - \log_9(4x) \;=\;\frac{1}{2}$

Then we have: . $\log_9\left(\frac{7x+6}{4x}\right) \;=\;\frac{1}{2}$

Exponentiate: . $\frac{7x+6}{4x} \;=\;9^{\frac{1}{2}} \quad\Rightarrow\quad \frac{7x+6}{4x} \;=\;3$

Therefore: . $7x+6 \:=\:12x\quad\Rightarrow\quad \boxed{x \:=\:\frac{6}{5}}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

**

Let $\log_3(7x+6) \:=\:P$

Then: . $3^P \:=\:7x+6$

Take logs, base 9: . $\log_9\left(3^P\right) \:=\;\log_9(7x+6) \quad\Rightarrow\quad P\!\cdot\!\log_9(3) \:=\:\log_9(7x+6)$

Since $\log_9(3) = \frac{1}{2}$, we have: . $\frac{1}{2}P \:=\:\log_9(7x+6) \quad\Rightarrow\quad P \:=\:2\!\cdot\!\log_9(7x+6)$

Therefore: . $\log_3(7x+6) \;=\;2\!\cdot\!\log_0(7x+6)$

• Mar 16th 2008, 04:04 AM
tyana21
yes you read it correctly.im a newbie so i have no idea how to write it out since i was in a rush.
• Mar 16th 2008, 04:40 AM
Moo
So don't bother with my message, because i read $\log_3 (7x+6) - \log_9(4x^2)=1$

:p
• Mar 16th 2008, 05:22 AM
tyana21
Quote:

Originally Posted by Moo
So don't bother with my message, because i read $\log_3 (7x+6) - \log_9(4x^2)=1$

:p

haha its okay i appreciate all the help