Kamren Franco

Answered

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)$4}^{3x}={8}^{x-1$

(a)

(b)

(c)

(d)

(e)

Answer & Explanation

fumanchuecf

Expert

2022-01-24Added 21 answers

I did only three first

a)${8}^{2-x}=2$

$8={2}^{3}$

${2}^{3(2-x)}=2$

So$3(2-x)=1$

$6-3x=1$

$3x=6-1$

$x=\frac{5}{3}$

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)$5}^{x}={25}^{x-2$

$25={5}^{2}$

$5}^{x}={\left(5\right)}^{2(x-2)$

So,$x=2(x-2)$

$x=2x-4$

$2x-x=4$

$x=4$

a)

So

b)

Thus,

c)

So,

Eleanor Shaffer

Expert

2022-01-25Added 16 answers

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.$4}^{3x}={8}^{x-1$

$4={2}^{2}$ , $8={2}^{3}$

$2}^{2\left(3x\right)}={2}^{3(x-1)$

$2\left(3x\right)=3(x-1)$

$6x=3x-3$

$6x-3x=-3$

$3x=-3$

$x=-1$

5.

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