How do you simplify the following?

1) $\frac{{25}^{2 t}}{{125}^{t}}$ 2) $\frac{{3}^{x + 3} - {3}^{x + 1}}{2} ^ 2$

Jun 27, 2018
• ${5}^{t}$

• ${3}^{x + 1} \times 2$

Explanation:

• ${25}^{2 t} / \left({125}^{t}\right)$

Simplify to the least term;

5^(2(2t))/(5^(3(t))

$\frac{{5}^{4 t}}{{5}^{3 t}}$

Recall;

${x}^{a} / {x}^{b} = {x}^{a - b}$

Therefore;

${5}^{4 t - 3 t}$

${5}^{t}$

• $\frac{{3}^{x + 3} - {3}^{x + 1}}{2} ^ 2$

Recall;

${x}^{a + b} = {x}^{a} \times {x}^{b}$

Therefore;

$\frac{{3}^{x} \times {3}^{3} - {3}^{x} \times {3}^{1}}{2} ^ 2$

Can also be;

$\frac{{3}^{x} \times {3}^{1 + 2} - {3}^{x} \times {3}^{1}}{2} ^ 2$

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

$\frac{\left({3}^{x} \times 3 \times 9\right) - \left({3}^{x} \times 3 \times 1\right)}{2} ^ 2$

Factorizing out the common terms

$\frac{{3}^{x} \times 3 \left(9 - 1\right)}{2} ^ 2$

$\frac{{3}^{x} \times 3 \left(8\right)}{2} ^ 2$

Remember;

${3}^{x} \times 3 = {3}^{x} \times {3}^{1} = {3}^{x + 1}$

Hence;

$\frac{{3}^{x + 1} \times 8}{2} ^ 2$

$\frac{{3}^{x + 1} \times {2}^{3}}{2} ^ 2$

${3}^{x + 1} \times {2}^{3 - 1}$

${3}^{x + 1} \times {2}^{1}$

${3}^{x + 1} \times 2$

Jun 27, 2018

1) => 5^t

2) => 2 (3 ^ (x + 1))

Explanation:

$1. {25}^{2 t} / {125}^{t}$

=> (cancel25^t * 25^t) / (cancel25^t * 5^t

$\implies \frac{{\cancel{5}}^{t} \cdot {5}^{t}}{\cancel{5}} ^ t = {5}^{t}$

$2. \frac{{3}^{x + 3} - {3}^{x + 1}}{2} ^ 2$

$\implies \frac{{3}^{x + 1} \cdot {3}^{2} - {3}^{x + 1}}{4}$

$\implies \frac{{3}^{x + 1} \cdot {\cancel{\left(9 - 1\right)}}^{\textcolor{red}{2}}}{\cancel{4}}$

$\implies 2 \left({3}^{x + 1}\right)$

Jun 27, 2018

${5}^{t} \text{ and } 6 {\left(3\right)}^{x}$

Explanation:

$\text{using the "color(blue)"laws of exponents}$

•color(white)(x)a^mxxa^n=a^((m+n))

•color(white)(x)a^m/a^n=a^((m-n))" and "(a^m)^n=a^(mn)

$\left(1\right)$

$\frac{{25}^{2 t}}{{125}^{t}}$

$= {\left({\left(5\right)}^{2}\right)}^{2 t} / {\left({\left(5\right)}^{3}\right)}^{t}$

$= {\left(5\right)}^{4 t} / {5}^{3 t} = {5}^{\left(4 t - 3 t\right)} = {5}^{t}$

$\left(2\right)$

$= \frac{{3}^{x} \left({3}^{3} - {3}^{1}\right)}{4}$

$= \frac{{3}^{x} \left(27 - 3\right)}{4}$

$= \frac{{\cancel{24}}^{6} \left({3}^{x}\right)}{\cancel{4}} ^ 1 = 6 {\left(3\right)}^{x}$

Jun 27, 2018
1. ${5}^{t}$

2. $2 \times {3}^{x + 1}$

Explanation:

For Question 1:

$\frac{{25}^{2 t}}{125} ^ t$ = ${\left({5}^{2}\right)}^{2 t} / {\left({5}^{3}\right)}^{t}$

By Exponential law:
${\left({x}^{a}\right)}^{b}$ = ${x}^{a \times b}$ =${x}^{a b}$

${x}^{a} / {x}^{b}$ =${x}^{a - b}$

So no we have:

$\frac{{25}^{2 t}}{125} ^ t$ = ${\left({5}^{2}\right)}^{2 t} / {\left({5}^{3}\right)}^{t}$ = $\frac{{5}^{2 \times 2 t}}{5} ^ \left(3 \times t\right)$ = ${5}^{4 t} / {5}^{3 t}$= ${5}^{4 t - 3 t}$ = ${5}^{t}$

Therefore:
$\frac{{25}^{2 t}}{125} ^ t$ = ${5}^{t}$

For Question 2:

First, lets simplify the numerator:

${3}^{x + 3} - {3}^{x + 1}$

${3}^{x + 1 + 2} - {3}^{x + 1}$

By Exponential Law:

${x}^{a + b}$ = ${x}^{a} . {x}^{b}$

So, ${3}^{x + 1 + 2}$ = ${3}^{2} \times {3}^{x + 1}$ = ${3}^{2} {.3}^{x + 1}$

We now have:

${3}^{2} {.3}^{x + 1} - {3}^{x + 1}$

Factor out the common term ${3}^{x + 1}$

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

${3}^{x + 1} . \left(9 - 1\right)$

$8 \times {3}^{x + 1}$

And now we have:

$\frac{{3}^{x + 1 + 2} - {3}^{x + 1}}{2} ^ 2$ = $\frac{8 \times {3}^{x + 1}}{4}$ = $2 \times {3}^{x + 1}$

$\frac{{3}^{x + 1 + 2} - {3}^{x + 1}}{2} ^ 2$ = $2 \times {3}^{x + 1}$