Part A: If (72)x = 1, what is the value of x? Explain your answer. (5 points) Part B: If (70)x = 1, what are the possible values of x? Explain your answer. (5 points)

Can you Help Me I NEED ANSWER ASAP!!!

see below

Step-by-step explanation:

Part A:  (72)^x = 1

Take the log base 72 of each side

log72(72^x) = log 72(1)

We know log a^b = b log a

x log72(72) = log72(1)

x =  log72(1)

x = 0

Part A:  (70)^x = 1

Take the log base 70 of each side

log70(70^x) = log70(1)

We know log a^b = b log a

x log70(70) = log70(1)

x =  log70(1)

x = 0

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}