Will someone help me with logarithm homework?

Question

1. Rewrite log base 7^343 = x in exponential form.

2. Solve 11^(n+1)= 13 for n.

3. Solve e^(3n)= 9 for n.

Answer 1

1. Log base 7 ³⁴³ = x in exponential form

      Take out log each side

         This is a exponential form.

 

 

 

 

Answer 2

3.

e^3n = 9

Take logarithm  to each side

loge³ⁿ = log9

The logarithm base is e

3n loge = log 9

Recall : loge = 1

log e = 1

3n = log9

3n = log 3²

3n = 2log3

Divide each side by 3

There fore n = 2/3 log3.

Answer 3

2.

11^{n + 1} = 13

Take logarithm to each side

log11 ^{n + 1} = log13

(n + 1)log11 = log13

Divide each side  by log11

(n + 1)log11 / log11 = log13 / log11

n+1 = log13 / log11

Substitute log13 =1.1139 and log11 = 1.043

There fore n + 1 = 1.1139 / 1.043

n + 1 = 1.06798

Subtract 1 from each side

n + 1 - 1 = 1.06798 - 1

n + 0 = 0.06798

n = 0.06798

n = 0.07 (approximately)

 

Answer 4

3.

e³ⁿ = 9

Take logarithm to each side

loge³ⁿ = log9

3nloge = log3²

3nloge = 2log3

Divide each side by 3loge

There fore n = 2/3log3 / loge

Substitute loge = 0.4343 and log3 = 0.4771

n = 2/3(0.4771 / 0.4343)

n = 2/3(1.099)

n = 0.73267 (approximately).

Answer 5

1) The logarthimic equation

The relationship

is equivalent to

Exponential form of is

Rewrite 343 as 7*7*7.

Solution x  = 3.