What is the answer to 1+3^{z^2-2}=5 ?

The answer to 1+3^{z^2-2}=5 is z=sqrt((2ln(2))/(ln(3))+2),z=-sqrt((2ln(2))/(ln(3))+2)

Solve for z: 1+3^{z^2-2}=5

The solution is z=sqrt((2ln(2))/(ln(3))+2),z=-sqrt((2ln(2))/(ln(3))+2)

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1 + 3^{z^{2} - 2} = 5

z = \frac{2 ln ( 2 )}{ln ( 3 )} + 2, z = - \frac{2 ln ( 2 )}{ln ( 3 )} + 2

Decimal Notation

z = 1.80606 \dots, z = - 1.80606 \dots

Solution steps

1 + 3^{z^{2} - 2} = 5

Move

1

to the right side

3^{z^{2} - 2} = 4

Apply exponent rules

(z^{2} - 2) ln (3) = ln (4)

Solve

(z^{2} - 2) ln (3) = ln (4) : z = \frac{2 ln ( 2 )}{ln ( 3 )} + 2, z = - \frac{2 ln ( 2 )}{ln ( 3 )} + 2

z = \frac{2 ln ( 2 )}{ln ( 3 )} + 2, z = - \frac{2 ln ( 2 )}{ln ( 3 )} + 2

3^{2 x + 1} + 9 = 28 \cdot 3^{x}

15 = 6 e^{0.07 k}

2^{x - 10} = x^{3}

e^{4 k} = \frac{3204}{2650}

3 3^{X} = 2

What is the answer to 1+3^{z^2-2}=5 ?
The answer to 1+3^{z^2-2}=5 is z=sqrt((2ln(2))/(ln(3))+2),z=-sqrt((2ln(2))/(ln(3))+2)
Solve for z: 1+3^{z^2-2}=5
The solution is z=sqrt((2ln(2))/(ln(3))+2),z=-sqrt((2ln(2))/(ln(3))+2)