1. ## 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)

2. for the first write it as
$\displaystyle (3^x)^2+3^x-2=0$

then set $\displaystyle u=3^x$ substitute in then solve the quadratic to find u then use the substitution and logs to find x

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

4. 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

5. 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)

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

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

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

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

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

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

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

$\displaystyle x\approx.234$

6. Hello, magentarita!

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

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

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

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

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

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

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

7. ## Thanks to all...Soroban

Thank you all. Soroban....I have seen you in other math forums.