Kamren Franco

2022-01-23

Please, solve for the values of x for each of the following exponential equations and inequalities.
(a) ${8}^{2-x}=2$
(b) ${\left(\frac{1}{2}\right)}^{x}<\left(\frac{1}{8}\right)$
(c) ${5}^{x}={25}^{x-2}$
(d) ${3}^{x+2}\ge 27$
(e) ${4}^{3x}={8}^{x-1}$

fumanchuecf

Expert

I did only three first
a)${8}^{2-x}=2$
$8={2}^{3}$
${2}^{3\left(2-x\right)}=2$
So $3\left(2-x\right)=1$
$6-3x=1$
$3x=6-1$
$x=\frac{5}{3}$
b)${\left(\frac{1}{2}\right)}^{x}<\left(\frac{1}{8}\right)$
$\frac{1}{{2}^{x}}<\frac{1}{8}$
$8<{2}^{x}$
${2}^{x}>8$
$8={2}^{3}$
${2}^{x}>{2}^{3}$
Thus, $x>3$
c)${5}^{x}={25}^{x-2}$
$25={5}^{2}$
${5}^{x}={\left(5\right)}^{2\left(x-2\right)}$
So, $x=2\left(x-2\right)$
$x=2x-4$
$2x-x=4$
$x=4$

Eleanor Shaffer

Expert

4. ${3}^{x+2}\ge 27$
$27={3}^{3}$
${3}^{x+2}\ge {3}^{3}$
$x+2\ge 3$
$x\ge 3-2$
$x\ge 1$
5. ${4}^{3x}={8}^{x-1}$
$4={2}^{2}$, $8={2}^{3}$
${2}^{2\left(3x\right)}={2}^{3\left(x-1\right)}$
$2\left(3x\right)=3\left(x-1\right)$
$6x=3x-3$
$6x-3x=-3$
$3x=-3$
$x=-1$

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