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belka [17]
1 year ago
10

Solve the following equation for X. Round your answer to four decimal places.

Mathematics
1 answer:
strojnjashka [21]1 year ago
4 0

ANSWER

x = 1.2226

EXPLANATION

To solve this equation we have to apply the property of the logarithm of the base,

\log _bb=1

Thus, we can apply the natural logarithm - whose base is e, to both sides of the equation,

\ln e^{4x}=\ln 133

Now we apply the property of the logarithm of a power,

\log a^b=b\log a

In our equation,

\begin{gathered} 4x\ln e^{}=\ln 133 \\ 4x^{}=\ln 133 \end{gathered}

Then divide both sides by 4 and solve,

\begin{gathered} \frac{4x^{}}{4}=\frac{\ln 133}{4} \\ x\approx1.2226 \end{gathered}

The solution to this equation is x = 1.2226, rounded to four decimal places.

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Given:

g(x)=4x^2+2x\\ \\f(x)=\int\limits^x_0 {g(t)} \, dt

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First, find f(x):

f(x)\\ \\=\int\limits^x_0 {g(t)} \, dt\\ \\=\int\limits^x_0 {(4t^2+2t)} \, dt\\ \\=\left(4\cdot \dfrac{t^3}{3}+2\cdot \dfrac{t^2}{2}\right)\big|\limits^x_0\\ \\=\left(\dfrac{4t^3}{3}+t^2\right)\big|\limits^x_0\\ \\= \left(\dfrac{4x^3}{3}+x^2\right)-\left(\dfrac{4\cdot 0^3}{3}+0^2\right)\\ \\=\dfrac{4x^3}{3}+x^2

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