Logarithmic and exponential function questions

**tyana21** · March 15th 2008, 05:55 AM

log3(7x+6)-log94x(squared)=1

ey-1=6/e
ln x2 +ey=18, find the value of (square root)ey(ln x3)

**Peritus** · March 15th 2008, 06:26 AM

$ \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} $

-----------------------------------------------------------

**Moo** · March 15th 2008, 06:47 AM

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.

**Soroban** · March 15th 2008, 09:48 AM

Hello, tyana21!

An I reading the first one correctly?

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

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