Respuesta :
27 = 9^-x + 5
Subtract 5 from both sides
22 = 9^-x
Log 22 = -x log 9
(Log22)/(log9) = -x
-(log22)/(log9)= x
-1.41 = x
Subtract 5 from both sides
22 = 9^-x
Log 22 = -x log 9
(Log22)/(log9) = -x
-(log22)/(log9)= x
-1.41 = x
1. Use negative power rule: x^-a=1/x^a
27=1/9^x+5
2. Subtract 5 from both sides
27-5=1/9^x
3. Simplify 27-5 to 22
22=1/9^x
4. Multiply both sides by 9^x
22*9^x=1
5. Divide both sides by 22
9^x=1/22
6. Use definition of common logarithm: b^a=x if and only if log(x)=a
x=log9(1/22)
7. Use change of base rule: logbx=(logan)/8logab)
x=(log1/22)/(log9)
8. Use power rule: logbx^c=clogbx log1/22 -> log(22^-1) -> -log22
x=(-log22)/(log9)
9. Use power rule
Use power rule: logbx^c=clogbx, log9 -> log3^2 -> 2 log3
x=(-log22)/(2log3)
10. Your answer is, and my wish of you having a nice day :D
x=-(log22)/(2log3)
x=(approx.)-1.41
27=1/9^x+5
2. Subtract 5 from both sides
27-5=1/9^x
3. Simplify 27-5 to 22
22=1/9^x
4. Multiply both sides by 9^x
22*9^x=1
5. Divide both sides by 22
9^x=1/22
6. Use definition of common logarithm: b^a=x if and only if log(x)=a
x=log9(1/22)
7. Use change of base rule: logbx=(logan)/8logab)
x=(log1/22)/(log9)
8. Use power rule: logbx^c=clogbx log1/22 -> log(22^-1) -> -log22
x=(-log22)/(log9)
9. Use power rule
Use power rule: logbx^c=clogbx, log9 -> log3^2 -> 2 log3
x=(-log22)/(2log3)
10. Your answer is, and my wish of you having a nice day :D
x=-(log22)/(2log3)
x=(approx.)-1.41