# Exponential Equations

• August 9th 2008, 06:43 AM
magentarita
Exponential Equations
Solve each exponential equation below. Express irrational solutions in exact form and as a decimal rounded to 3 decimal places.

(1) 3^(2x) + 3^(x) - 2 = 0

(2) 2^(x + 1) = 5^(1 - 2x)

• August 9th 2008, 09:08 AM
thelostchild
for the first write it as
$(3^x)^2+3^x-2=0$

then set $u=3^x$ substitute in then solve the quadratic to find u then use the substitution and logs to find x
• August 9th 2008, 01:43 PM
magentarita
You said to let u = 3^(x). Of course, I can select any letter of choice to represent 3^(x), right?
• August 9th 2008, 02:26 PM
thelostchild
Quote:

Originally Posted by magentarita
You said to let u = 3^(x). Of course, I can select any letter of choice to represent 3^(x), right?

of course you can choose any letter under the sun :)
• August 9th 2008, 02:27 PM
masters
Quote:

Originally Posted by magentarita
You said to let u = 3^(x). Of course, I can select any letter of choice to represent 3^(x), right?

Correct, any letter will do.

(2)

$2^{x+1}=5^{1-2x}$

$\log$
$2^{x+1}=\log5^{1-2x}$

$(x+1)\log2=(1-2x)\log5$

$x\log2+\log2=\log5-2x\log5$

$x\log2+2x\log5=\log5-\log2$

$x(\log2+2\log5)=\log5-\log2$

$x=\frac{\log5-\log2}{2\log5+\log2}$

$x\approx.234$
• August 9th 2008, 07:29 PM
Soroban
Hello, magentarita!

There are a number of ways to solve #2 . . .

Quote:

Solve the following equation. Express irrational solutions
in exact form and as a decimal rounded to 3 decimal places.

$2)\;\;2^{x + 1} \;=\; 5^{1 - 2x}$

We have: . $2^x\cdot2^1 \;=\;5^1\cdot5^{-2x} \quad\Rightarrow\quad 2^x\cdot5^{2x} \;=\;\frac{5}{2} \quad\Rightarrow\quad 2^x\cdot(5^2)^x\;=\;\frac{5}{2}$

. . $=\; 2^x\cdot25^x \;=\;\frac{5}{2} \quad\Rightarrow\quad (2\cdot25)^x \;=\;\frac{5}{2} \quad\Rightarrow\quad 50^x \;=\;2.5$

Take logs: . $\ln(50^x) \;=\;\ln(2.5) \quad\Rightarrow\quad x\cdot\ln(50) \;=\;\ln(2.5)$

. . Therefore: . $x \;=\;\boxed{\frac{\ln(2.5)}{\ln(50)}} \;\approx\;\boxed{0.234}$

• August 10th 2008, 06:20 AM
magentarita
Thanks to all...Soroban
Thank you all. Soroban....I have seen you in other math forums.