What is the answer to solvefor d,5ln(d^2+yx)=7 ?

The answer to solvefor d,5ln(d^2+yx)=7 is d=sqrt(e^{7/5)-yx},d=-sqrt(e^{7/5)-yx}

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solvefor d,5ln(d^2+yx)=7

solve for

d, 5 ln (d^{2} + y x) = 7

d = e^{\frac{7}{5}} - y x, d = - e^{\frac{7}{5}} - y x

Solution steps

5 ln (d^{2} + y x) = 7

Divide both sides by

5

ln (d^{2} + y x) = \frac{7}{5}

Apply log rules

d^{2} + y x = e^{\frac{7}{5}}

Solve

d^{2} + y x = e^{\frac{7}{5}} : d = e^{\frac{7}{5}} - y x, d = - e^{\frac{7}{5}} - y x

d = e^{\frac{7}{5}} - y x, d = - e^{\frac{7}{5}} - y x

2 lo g_{4} (x) - lo g_{4} (x + 3) = 1

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

- ln (1 - x) = 0

lo g_{2} (c) = 4 + lo g_{2} (c - 3)

lo g_{0.5} (x) = 5

What is the answer to solvefor d,5ln(d^2+yx)=7 ?
The answer to solvefor d,5ln(d^2+yx)=7 is d=sqrt(e^{7/5)-yx},d=-sqrt(e^{7/5)-yx}