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mote1985 [20]
3 years ago
12

How do you use the properties of exponents and logarithms to rewrite functions in equivalent forms and solve equations

Mathematics
1 answer:
Leona [35]3 years ago
7 0
Exponential:
 
It is called the exponential function of base a, to that whose generic form is f (x) = a ^ x, being a positive number other than 1.
 Every exponential function of the form f (x) = a^x, complies with the followingProperties:
 1. The function applied to the zero value is always equal to 1: f (0) = a ^ 0  = 1
 2. The exponential function of 1 is always equal to the base: f (1) = a ^ 1  = a.
 3. The exponential function of a sum of values is equal to the product of the application of said function on each value separately.
 f (m + n) = a ^ (m + n) = a ^ m · a ^ n
 = f (m) · f (n).
 4. The exponential function of a subtraction is equal to the quotient of its application to the minuend divided by the application to the subtrahend:
 f (p - q) = a ^ (p - q) = a ^ p / a ^ q
 Logarithm:
 
In the loga (b), a is called the base of the logarithm and b is called an argument, with a and b positive.
 Therefore, the definition of logarithm is:
 loga b = n ---> a ^ n = b (a> 0, b> 0)
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Answer:

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Step-by-step explanation:

The given differential equation is

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Taking out common factors.

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Using variable separable method, we get

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Integrate both sides.

\int\frac{dz}{5z+2}=\int e^{\sin t}\cos t dt

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First solve LHS,

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Substitute 5z+2=u

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dz=\frac{1}{5}du

I_1=\frac{1}{5}\int\frac{du}{u}

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I_1=\frac{1}{5}ln|5z+2|+C_1

Now, solve RHS,

I_2=\int e^{\sin t}\cos t dt

Substitute \sin t=v,

\cos t\frac{dt}{dv}=1

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I_2=e^{\sin t}+C_2

Subtitle the values of I₁ and I₂ in equation (1).

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\frac{1}{5}ln|5z+2|=e^{\sin t}+C_2-C_1

\frac{1}{5}ln|5z+2|=e^{\sin t}+C

Therefore the general solution of given differential equation is \frac{1}{5}ln|5z+2|=e^{\sin t}+C.

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