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.

Respuesta :

Answer:

Part a) [tex]n=-8.017176[/tex]

Part b) [tex]n=9.700465[/tex]

Step-by-step explanation:

Part a) we have

[tex]9^{(n+10)}+3=81[/tex]

Solve for n

Subtract 3 both sides

[tex]9^{(n+10)}+3-3=81-3[/tex]

[tex]9^{(n+10)}=78[/tex]

take log both sides

[tex]log[9^{(n+10)}]=log[78][/tex]

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

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

[tex]n=\frac{log(78)}{log(9)}-10[/tex] ----> using a calculator

[tex]n=-8.017176[/tex]

Part b) we have

[tex]11^{(n-8)}-5=54[/tex]

Solve for n

Adds 5 both sides

[tex]11^{(n-8)}-5+5=54+5[/tex]

[tex]11^{(n-8)}=59[/tex]

take log both sides

[tex]log[11^{(n-8)}]=log[59][/tex]

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

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

[tex]n=\frac{log(59)}{log(11)}+8[/tex] ----> using a calculator

[tex]n=9.700465[/tex]