What is the answer to solvefor x,ln(x/(x_{0)})=u(t-t_{0}) ?

The answer to solvefor x,ln(x/(x_{0)})=u(t-t_{0}) is x=e^{u(t-t_{0})}x_{0}

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solvefor x,ln(x/(x_{0)})=u(t-t_{0})

solve for

x, ln (\frac{x}{x _{0}}) = u (t - t_{0})

x = e^{u (t - t_{0})} x_{0}

Solution steps

ln (\frac{x}{x _{0}}) = u (t - t_{0})

Apply log rules

\frac{x}{x _{0}} = e^{u (t - t_{0})}

Multiply both sides by

x_{0}

\frac{x}{x _{0}} x_{0} = e^{u (t - t_{0})} x_{0}

Simplify

x = e^{u (t - t_{0})} x_{0}

2.8 = - lo g_{10} (h^{30})

2 ln (x) - 2 ln (x + 2) = 4 ln (x^{2})

ln (x^{2} - x - 1) = 0

lo g_{3} (2 x^{2} + 5 x + 4) - 7 = 2

\frac{lo g _{10} ( 6 - x ^{2} )}{lo g _{10} ( 3 x - 4 )} = 2

What is the answer to solvefor x,ln(x/(x_{0)})=u(t-t_{0}) ?
The answer to solvefor x,ln(x/(x_{0)})=u(t-t_{0}) is x=e^{u(t-t_{0})}x_{0}