How do you simplify the following?

1) (25^(2t))/(125^t)

2) (3^(x+3)-3^(x+1))/2^2

4 Answers
Jun 27, 2018
  • 5^t

  • 3^(x + 1) xx 2

Explanation:

  • 25^(2t)/(125^t)

Simplify to the least term;

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

(5^(4t))/(5^(3t))

Recall;

x^a/x^b = x^(a - b)

Therefore;

5^(4t - 3t)

5^t

  • (3^(x + 3) - 3^(x + 1))/2^2

Recall;

x^(a + b) = x^a xx x^b

Therefore;

(3^x xx 3^3 - 3^x xx 3^1)/2^2

Can also be;

(3^x xx 3^(1 + 2) - 3^x xx 3^1)/2^2

((3^x xx 3^1 xx 3^2) - (3^x xx 3^1))/2^2

((3^x xx 3 xx 9) - (3^x xx 3 xx 1))/2^2

Factorizing out the common terms

(3^x xx 3(9 - 1))/2^2

(3^x xx 3(8))/2^2

Remember;

3^x xx 3 = 3^x xx 3^1 = 3^(x + 1)

Hence;

(3^(x + 1) xx 8)/2^2

(3^(x + 1) xx 2^3)/2^2

3^(x + 1) xx 2^(3 - 1)

3^(x + 1) xx 2^1

3^(x + 1) xx 2

Jun 27, 2018

1) => 5^t

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

Explanation:

https://www.youtube.com/watch?v=ARLS2TmFT94https://www.youtube.com/watch?v=ARLS2TmFT94

1. 25^(2t) / 125^t

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

=> (cancel5^t * 5^t) / cancel5^t = 5^t

2. (3^(x + 3) - 3^(x + 1)) / 2^2

=> (3 ^(x + 1) * 3^2 - 3^(x + 1))/4

=>(3^(x + 1) * cancel((9 - 1))^color(red)(2)) / cancel4

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

Jun 27, 2018

5^t" and "6(3)^x

Explanation:

"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)

(1)

(25^(2t))/(125^t)

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

=(5)^(4t)/5^(3t)=5^((4t-3t))=5^t

(2)

=(3^x(3^3-3^1))/4

=(3^x(27-3))/4

=(cancel(24)^6(3^x))/cancel(4)^1=6(3)^x

Jun 27, 2018
  1. 5^t

  2. 2 xx 3^(x+1)

Explanation:

For Question 1:

(25^(2t))/125^t = (5^2)^(2t)/(5^3)^t

By Exponential law:
(x^a)^b = x^(a xx b) = x^(ab)

x^a/x^b =x^(a-b)

So no we have:

(25^(2t))/125^t = (5^2)^(2t)/(5^3)^t = (5^(2 xx 2t))/5^(3 xx t) = 5^(4t)/5^(3t)= 5^(4t-3t) = 5^t

Therefore:
(25^(2t))/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 xx 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).(3^2 - 1)

3^(x+1).(9-1)

8 xx 3^(x+1)

And now we have:

(3^(x+1+2) - 3^(x+1))/2^2 = (8 xx 3^(x+1))/4 = 2 xx 3^(x+1)

(3^(x+1+2) - 3^(x+1))/2^2 = 2 xx 3^(x+1)