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)
Correct, any letter will do.
(2)
$\displaystyle 2^{x+1}=5^{1-2x}$
$\displaystyle \log$$\displaystyle 2^{x+1}=\log5^{1-2x}$
$\displaystyle (x+1)\log2=(1-2x)\log5$
$\displaystyle x\log2+\log2=\log5-2x\log5$
$\displaystyle x\log2+2x\log5=\log5-\log2$
$\displaystyle x(\log2+2\log5)=\log5-\log2$
$\displaystyle x=\frac{\log5-\log2}{2\log5+\log2}$
$\displaystyle x\approx.234$
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: .$\displaystyle x \;=\;\boxed{\frac{\ln(2.5)}{\ln(50)}} \;\approx\;\boxed{0.234}$