exponential equation

**icefirez** · February 6th 2010, 01:52 AM

i need to solve this but it's kind of difficult when i try it by myselfe

hope u explain me how u solved than...

**HallsofIvy** · February 6th 2010, 02:06 AM

First step, combine the like bases:
$2^{\frac{x- 2}{3}}+ 2^{\frac{x+1}{3}}= 3^{\frac{x-1}{2}}- 3^{\frac{x-3}{2}}$

Now use laws of exponents to take out "like parts". $\frac{x- 2}{3}= \frac{x}{3}- \frac{2}{3}$ and $\frac{x+1}{3}= \frac{x}{3}+ \frac{1}{3}$ so

$2^{\frac{x}{3}- \frac{2}{3}}+ 2^{\frac{x}{3}- \frac{1}{3}}$ $= 2^{\frac{x}{3}}\left(2^{-\frac{2}{3}}+ 2^{\frac{1}{3}}\right)$ where the value inside the parentheses is a number you can calculate directly. Do the same thing on the right so you get something like $A2^{\frac{x}{3}}= B3^{\frac{x}{2}}$ where "A" and "B" are specific numbers. Finally, take logarithms of both sides.

Added: the "A" and "B" are $A= 2^{-\frac{2}{3}}+ 2^{\frac{1}{3}}$ which, according to my calculator is 1.8900 (to four decimal places) and [tex]B= 3^{-1/2}- 3^{-3/2}= 0.3849.

Your equation is $(1.8900)2^{x/3}= (0.3849)3^{x/2}$ taking logarithms of both sides, log(1.8900)+ (log(2)/3)x= log(0.3849)+ (log(3)/2)x. You can use either common or natural logarithms and get the correct answer.

**icefirez** · February 6th 2010, 02:17 AM

actually i still cant get this...i'm new to exponential equations...i thought u will go to the end of this task...and we didnt learned logarithms, but i just readed some lessons in internet about logarithms...anyway thanks i will keep practising and practising over over again ...probably be like you

in math i hope soo ...

**Soroban** · February 6th 2010, 09:15 AM

Hello, icefirez!

Okay, here it is in baby steps . . .

$\Large 2^{\frac{x-2}{3}} + 3^{\frac{x-3}{2}} \;=\;3^{\frac{x-1}{2}} - 2^{\frac{x+1}{3}}$

We have: . $2^{\frac{x-2}{3}} + 2^{\frac{x+1}{3}} \;=\;3^{\frac{x-1}{2}} - 3^{\frac{x-3}{2}}$

Factor: . $2^{\frac{x=2}{3}}(1 + 2) \;=\;3^{\frac{x-3}{2}}(3 - 1) \quad\Rightarrow\quad 2^{\frac{x-2}{3}}\cdot3 \;=\;3^{\frac{x-3}{2}}\cdot2$

. . . . . $\frac{2^{\frac{x-2}{3}}} {2} \;=\;\frac{3^{\frac{x-2}{2}}}{3} \quad\Rightarrow\quad 2^{\frac{x-5}{3}} \;=\;3^{\frac{x-4}{2}}$

Take logs: . $\ln\left(2^{\frac{x-5}{3}}\right) \;=\;\ln\left(3^{\frac{x-4}{2}}\right) \quad\Rightarrow\quad \frac{x-5}{3}\cdot\ln 2 \;=\;\frac{x-4}{2}\cdot\ln 3$

Multiply by 6: . $2(x-5)\ln 2 \;=\;3(x-4)\ln 3 \quad\Rightarrow\quad 2x\ln 2 - 10\ln 2 \;=\;3x\ln 3 - 12\ln 3$

. . . . . . . $3x\ln 3 - 2x\ln 2 \;=\;12\ln 3 - 10\ln 2 \quad\Rightarrow\quad x(3\ln3 - 2\ln 2) \;=\;12\ln3 - 10\ln 2$

Therefore: . $x \;=\;\frac{12\ln3 - 10\ln2}{3\ln3 - 2\ln2}$

But check my work . . . please!
.

**icefirez** · February 6th 2010, 10:46 AM

thanks it was kind of that i solved with my profesor in class...it's much easier now...but ok im understanding this ...thanks you

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