What is the answer to (a+1)^{(a+1)}=9^4 ?

The answer to (a+1)^{(a+1)}=9^4 is a=(8ln(3))/(\W_{0)(8ln(3))}-1

Solve for a: (a+1)^{(a+1)}=9^4

The solution is a=(8ln(3))/(\W_{0)(8ln(3))}-1

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(a + 1)^{(a + 1)} = 9^{4}

a = \frac{8 ln ( 3 )}{W _{0} ( 8 ln ( 3 ) )} - 1

Decimal Notation

a = 4.28126 \dots

Solution steps

(a + 1)^{(a + 1)} = 9^{4}

Prepare

(a + 1)^{(a + 1)} = 9^{4}

for Lambert form:

(a + 1) e^{- \frac{8 l n ( 3 )}{a + 1}} = 1

Rewrite the equation with

\frac{8 ln ( 3 )}{a + 1} = u

and

a = \frac{8 ln ( 3 )}{u} - 1

((\frac{8 ln ( 3 )}{u} - 1) + 1) e^{- u} = 1

Rewrite

((\frac{8 ln ( 3 )}{u} - 1) + 1) e^{- u} = 1

in Lambert form:

u e^{u} = 8 ln (3)

Solve

u e^{u} = 8 ln (3) : u = W_{0} (8 ln (3))

u = W_{0} (8 ln (3))

Substitute back

u = \frac{8 ln ( 3 )}{a + 1},

solve for

a

Solve

\frac{8 ln ( 3 )}{a + 1} = W_{0} (8 ln (3)) : a = \frac{8 ln ( 3 )}{W _{0} ( 8 ln ( 3 ) )} - 1

a = \frac{8 ln ( 3 )}{W _{0} ( 8 ln ( 3 ) )} - 1

6^{x} = a^{\frac{x}{4}}

0.1 = 1 0^{x}

e^{2 x} = - x + 2

7^{3 x - 5} = 49

e^{2 x} = 4 e^{x} - 4

What is the answer to (a+1)^{(a+1)}=9^4 ?
The answer to (a+1)^{(a+1)}=9^4 is a=(8ln(3))/(\W_{0)(8ln(3))}-1
Solve for a: (a+1)^{(a+1)}=9^4
The solution is a=(8ln(3))/(\W_{0)(8ln(3))}-1