Solve single exponent

Question

Let m(x) = x 1

6 , n(x) = x3, and p(x) = x4.

(a) Write m(n(x)) as a function of x with a single exponent.

(b) Write m(n(x)p(x)) as a function of x with a single exponent.

(c) For what values of x is m(p(x)) = x 2

3 ? Explain.

(d) For what values of x is 3pn(x) = x? Explain.

Answer 1

m(x)=x^16

n(x) = x^3

p(x) = x^4

m(n(x)) = m(x^3) = (x^3)^16 = x^48

m(n(x)p(x)) = m(x^3x^4) = m(x^7) = (x^7)^16 = x^112

m(p(x)) = x^2 or x^23

m(x^4) = (x^4)^16 =x^64 = x^2 or x^23 when x=1. one to any power = 1

x^64=x^2

x^62=1

x = the 62nd root of 1. 62 possible roots, including 1 and -1. 60 roots are imaginary,

not sure what pn(x) means.

if you mean p(x)n(x) then

3p(x)n(x) = 3x^4x^3=3x^7

3x^7 =x

divide both sides by 3x

3x^6 =1

which only = x when 3x^6=1

x^6=1/3

x = the 6th root of 1/3, 6 possible roots

2 real, 4 imaginary, imaginary come in conjgate pairs,

be careful multiplying or dividing by a variable, as you may introduce extraneous roots

if you mean x^23 then

x^64=x^23

x^41 = 1

41 possible roots, including 1, most all imaginary

graph the equation, where it intersects the x axis gives the values of x that are solutions