What is the answer to solvefor x=ln(y^2)-0.7,y ?

The answer to solvefor x=ln(y^2)-0.7,y is y=sqrt(e^{x+0.7)},y=-sqrt(e^{x+0.7)}

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solvefor x=ln(y^2)-0.7,y

solve for

x = ln (y^{2}) - 0.7, y

y = e^{x + 0.7}, y = - e^{x + 0.7}

Solution steps

x = ln (y^{2}) - 0.7

Switch sides

ln (y^{2}) - 0.7 = x

Move

0.7

to the right side

ln (y^{2}) = x + 0.7

Apply log rules

y^{2} = e^{x + 0.7}

Solve

y^{2} = e^{x + 0.7} : y = e^{x + 0.7}, y = - e^{x + 0.7}

y = e^{x + 0.7}, y = - e^{x + 0.7}

Verify Solutions:

y = e^{x + 0.7}

True

, y = - e^{x + 0.7}

True

The solutions are

y = e^{x + 0.7}, y = - e^{x + 0.7}

ln (x^{3} - 4) = 6

lo g_{10} (x + 9) = 2

lo g_{10} (x) = lo g_{10} (x)

lo g_{0.1} (x^{2} + 3 x) = - 1

3 x^{2} ln (x) + x^{2} = 0

What is the answer to solvefor x=ln(y^2)-0.7,y ?
The answer to solvefor x=ln(y^2)-0.7,y is y=sqrt(e^{x+0.7)},y=-sqrt(e^{x+0.7)}