Log2(x) + log2(2x) + log2(3x) + log2(36) = 6

x = 2/3

Step-by-step explanation:

log₂(x) + log₂(2x) + log₂(3x) + log₂(36) = 6

According to the following properties of logarithm, we know that:

logₐ(x.y) = logₐ(x) + logₐ(y)

logₐ(x) = n

logₐ(x) = logₐ(y) ⇔ x = y

Applying this properties to the situation:

log₂(x.2x.3x.36) = log₂64

log₂(216x³) = log₂64

216x³ = 64

x³ = 64/216

x³ = 8/27

x = 2/3

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}