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Base change formula of logarithms Calculator

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1

Solved example of evaluate logarithms

$\frac{dy}{dx}=e^{3x+2y}$
2

Apply the property of the product of two powers of the same base in reverse: $a^{m+n}=a^m\cdot a^n$

$\frac{dy}{dx}=e^{3x}e^{2y}$
3

Group the terms of the differential equation. Move the terms of the $y$ variable to the left side, and the terms of the $x$ variable to the right side

$\frac{1}{e^{2y}}dy=e^{3x}dx$
4

Integrate both sides, the left side with respect to $y$, and the right side with respect to $x$

$\int\frac{1}{e^{2y}}dy=\int e^{3x}dx$

We can solve the integral $\int\frac{1}{e^{2y}}dy$ by applying integration by substitution method (also called U-Substitution). First, we must identify a part of the integral with a new variable, which when substituted makes the integral easier. We see that $e^{2y}$ it's a good candidate for substitution. Let's define a variable $u$ and assign it to the choosen part

$u=e^{2y}$

Now, in order to rewrite $dy$ in terms of $du$, we need to find the derivative of $u$. We need to calculate $du$, we can do that by deriving the equation above

$du=2e^{2y}dy$

Isolate $dy$ in the previous equation

$\frac{du}{2e^{2y}}=dy$

Substituting $u$ and $dy$ in the integral and simplify

$\frac{1}{2}\int\frac{1}{u^2}du$

Rewrite the exponent using the power rule $\frac{a^m}{a^n}=a^{m-n}$, where in this case $m=0$

$\frac{1}{2}\int u^{-2}du$

Apply the power rule for integration, $\displaystyle\int x^n dx=\frac{x^{n+1}}{n+1}$, where $n$ represents a constant function

$-\frac{1}{2}u^{-1}$

Substitute $u$ back with the value that we assigned to it: $e^{2y}$

$-\frac{1}{2}e^{-2y}$
5

Solve the integral $\int\frac{1}{e^{2y}}dy$ and replace the result in the differential equation

$-\frac{1}{2}e^{-2y}=\int e^{3x}dx$

We can solve the integral $\int e^{3x}dx$ by applying integration by substitution method (also called U-Substitution). First, we must identify a part of the integral with a new variable, which when substituted makes the integral easier. We see that $3x$ it's a good candidate for substitution. Let's define a variable $u$ and assign it to the choosen part

$u=3x$

Now, in order to rewrite $dx$ in terms of $du$, we need to find the derivative of $u$. We need to calculate $du$, we can do that by deriving the equation above

$du=3dx$

Isolate $dx$ in the previous equation

$\frac{du}{3}=dx$

Substituting $u$ and $dx$ in the integral and simplify

$\int\frac{e^u}{3}du$

Take the constant $\frac{1}{3}$ out of the integral

$\frac{1}{3}\int e^udu$

The integral of the exponential function is given by the following formula $\displaystyle \int a^xdx=\frac{a^x}{\ln(a)}$, where $a > 0$ and $a \neq 1$

$\frac{1}{3}e^u$

Substitute $u$ back with the value that we assigned to it: $3x$

$\frac{1}{3}e^{3x}$
6

Solve the integral $\int e^{3x}dx$ and replace the result in the differential equation

$-\frac{1}{2}e^{-2y}=\frac{1}{3}e^{3x}$
7

As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration $C$

$-\frac{1}{2}e^{-2y}=\frac{1}{3}e^{3x}+C_0$
8

Eliminate the $-\frac{1}{2}$ from the left, multiplying both sides of the equation by $$

$e^{-2y}=-2\left(\frac{1}{3}e^{3x}+C_0\right)$
9

Solve the product $-2\left(\frac{1}{3}e^{3x}+C_0\right)$

$e^{-2y}=-\frac{2}{3}e^{3x}-2C_0$
10

We can rename $-2C_0$ as other constant

$e^{-2y}=-\frac{2}{3}e^{3x}+C_1$
11

We can take out the unknown from the exponent by applying natural logarithm to both sides of the equation

$\ln\left(e^{-2y}\right)=\ln\left(-\frac{2}{3}e^{3x}+C_1\right)$
12

Using the power rule of logarithms: $\log_a(x^n)=n\cdot\log_a(x)$

$-2y\ln\left(e\right)=\ln\left(-\frac{2}{3}e^{3x}+C_1\right)$
13

Calculating the natural logarithm of $e$

$-2\cdot 1y=\ln\left(-\frac{2}{3}e^{3x}+C_1\right)$
14

Any expression multiplied by $1$ is equal to itself

$-2y=\ln\left(-\frac{2}{3}e^{3x}+C_1\right)$
15

Divide both sides of the equation by $-2$

$y=\frac{\ln\left(-\frac{2}{3}e^{3x}+C_1\right)}{-2}$

Final Answer

$y=\frac{\ln\left(-\frac{2}{3}e^{3x}+C_1\right)}{-2}$

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