How do you simplify the following?

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

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

4 Answers
Jumbotron
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#

sankarankalyanam
Jun 27, 2018

#1) => 5^t#

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

Explanation:

https://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))#

Jim G.
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#

Bhavin B.
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)#

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