Question

Help with algebra 2?

Answer 1

Sure, what is the question?

Answer 2

I think that your answer is correct!

Answer 3

@crustymcbeth

Answer 4

assume :-
u=
x
b, then x=u1b

 and dx=1b

du
x
b−1

∫
∞
0
x
a
1+
x
b



=
1
b

∫
∞
0
x
a
(1+u)
x
b−1

du
=
1
b

∫
∞
0
x
a−b+1
(1+u)

du
=
1
b

∫
∞
0
u
a+1
b

−1
(1+u)

du
 let m=
a+1
b

 and n+m=1 then n=1−
a+1
b



1
b

∫
∞
0
u
a+1
b

−1
(1+u)

du





=
1
b

∫
∞
0
u
m−1
(1+u
)
n+m

du
=
1
b

β(m,n)
=
1
b

β(
a+1
b

,1−
a+1
b

)
=
1
b

Γ(
a+1
b

)Γ(1−
a+1
b

)
Γ(
a+1
b

+1−
a+1
b

)

=
1
b

Γ(
a+1
b

)Γ(1−
a+1
b

)
Γ(1)

=1b

π
sin(π(a+1b))

we have to go in the complex plane,namely I set:
x
(b/2)=z
so your integral becomes:
2b
∫
0
∞
z
2a/b+2/b−1
1+
z2dz
now that integral is equal to:
2π/2

(Resf(i)+Resf(−i))
where Res is the residual value of f(z) and f(z) is our integrand above.
after a standard calculation, we get:
πb4

(e)
iπ(a+1)/b−e
−iπ(a+1)/b)
=πb4

∗2isin(πa+1b)
and taking the real part we can write:
πb2

∗sin(πa+1b)

Answer 5

please you have to use this properties of logarithms:
\[n*\log _{5}x=\log _{5}(x)^{n}\]
where n can be 0, 1, 2, 3,... or 0, -1, -2, -3,...as well
\[\log _{5}x+\log _{5}y=\log _{5}(x+y)\]
and this:
\[\frac{ 1 }{ n }\log _{5}x=\log _{5}\sqrt[n]{x}\]
where n, can be equal to 1,2,3,4,...

Answer 6

oops... these properties...

Answer 7

furthermore, this property can be useful:
\[\log _{5}x-\log _{5}y=\log _{5}\left( \frac{ x }{ y } \right)\]

Answer 8

please, try to apply those properties above, in order to simplify your expression

Answer 9

i still dont know what to do

Answer 10

let me briefly post the rules:

\(\Large\color{royalblue}{ c\cdot\log_a(b)= \log_a (b^c) }\) (exponent)

\(\Large\color{blueviolet}{\log_a(d)+\log_a(f)= \log_a (d \cdot f) }\) (addition)

\(\Large\color{royalblue}{\log_a(d)-\log_a(f)= \log_a (d \div f) }\) (subtraction)

Answer 11

u have to know the formulas

Answer 12

@SolomonZelman has mentioned them

Answer 13

apply the rules respectively.... and you should be alright