Question

What is the constant in

a) 9^n+10 + 3 = 81

b) 11^n−8 − 5 = 54

Edit: Never mind I figured it out this is a free 5-10 points answer whatever you want I won't delete it.
Good Luck.

Answer 1

Part a)
`n=-8.017176`

Part b)
`n=9.700465`

Explanation:

Part a) we have

`9^((n+10))+3=81`

Solve for n

Subtract 3 both sides

`9^((n+10))+3-3=81-3`

`9^((n+10))=78`

take log both sides

`log[9^((n+10))]=log[78]`

`(n+10)log[9]=log[78]`

`(n+10)=log[78]/log[9]`

`n=(log(78))/(log(9))-10` ----> using a calculator

`n=-8.017176`

Part b) we have

`11^((n-8))-5=54`

Solve for n

Adds 5 both sides

`11^((n-8))-5+5=54+5`

`11^((n-8))=59`

take log both sides

`log[11^((n-8))]=log[59]`

`(n-8)log[11]=log[59]`

`(n-8)=log[59]/log[11]`

`n=(log(59))/(log(11))+8` ----> using a calculator

`n=9.700465`