What is the answer to solvefor x,z(x)=xln(1+2x) ?

The answer to solvefor x,z(x)=xln(1+2x) is x=0,x=(e^z-1)/2

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solvefor x,z(x)=xln(1+2x)

solve for

x, z (x) = x ln (1 + 2 x)

x = 0, x = \frac{e ^{z} - 1}{2}

Solution steps

z (x) = x ln (1 + 2 x)

Switch sides

x ln (1 + 2 x) = z x

Move

z x

to the left side

x ln (1 + 2 x) - z x = 0

Factor

x ln (1 + 2 x) - z x : x (ln (1 + 2 x) - z)

x (ln (1 + 2 x) - z) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x = 0 or ln (1 + 2 x) - z = 0

Solve

ln (1 + 2 x) - z = 0 : x = \frac{e ^{z} - 1}{2}

Verify Solutions:

x = 0

True

, x = \frac{e ^{z} - 1}{2}

True

The solutions are

x = 0, x = \frac{e ^{z} - 1}{2}

2 lo g_{4} (x) - 1 = 0

- lo g_{10} (x) = 0.209

2 ln (x - 3) + ln (x + 7) = 3 ln (x)

y, 2 ln (y) = x^{2} + c

lo g_{10} (x) = - 2.6

What is the answer to solvefor x,z(x)=xln(1+2x) ?
The answer to solvefor x,z(x)=xln(1+2x) is x=0,x=(e^z-1)/2