Part A: If (6^2)^X = 1, what is the value of x? Explain your answer. (5 points)
Part B: If (6^9)^x = 1, what are the possible values of x? Explain your answer.

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absor201 absor201 · Answer 1 · 2019-06-07T06:00:12+02:00

Answer:

Part A: X=0

Part B: x=0

Step-by-step explanation:

Part A

(6^2)^X = 1

Applying the exponent rule: [tex](a^b)^c = a^{bc}[/tex]

So, our equation will become:

[tex]6^{2X} = 1[/tex]

We know if f(x) = g(x) then ln(f(x))= ln(g(x))

SO, taking natural logarithm ln on both sides and solving.

[tex]ln(6^{2X}) =ln(1)[/tex]

We know,[tex]log(a^b) = b.loga[/tex] Applying the rule,

[tex]2Xln6 =ln(1)\\We\,\,know\,\,ln(1)=0\\2Xln6 =0\\Solving:\\X=0[/tex]

Part B

(6^9)^x = 1

Applying the exponent rule: [tex](a^b)^c = a^{bc}[/tex]

So, our equation will become:

[tex]6^{9x} = 1[/tex]

We know if f(x) = g(x) then ln(f(x))= ln(g(x))

SO, taking natural logarithm ln on both sides and solving.

[tex]ln(6^{9x}) =ln(1)[/tex]

We know,[tex]ln(a^b) = b.lna[/tex] Applying the rule,

[tex]9xln6 =ln(1)\\We\,\,know\,\,ln(1)=0\\9xln6 =0\\Solving:\\x=0[/tex]