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Equations and functions
The function properties, like increasing and decreasing, extreme values, convexity or concavity are tools that can be well used while solving equations and inequalities in certain exercises. The following exercises give a few examples for this.
Example 1
Let us solve the equation \latex{\log_2 x =3-x} on the set of positive real numbers.
Solution
Let us again define two functions (Figure 12), which map from the set of positive real numbers to the set of real numbers:
\latex{f:\R^+\rightarrow\R, \;\;f(x)=\log_2x} and \latex{g:\R^+\rightarrow\R,\;\;g(x)=3-x}.
The function \latex{ f } is strictly increasing on its domain, the function \latex{ g } is strictly decreasing, thus their values can be equal at one place at most: at the place \latex{x = 2 f(2) = g(2) = 1}. So the single root of the equation is \latex{x = 2}.
\latex{y=3-x}
\latex{y=\log_2x}
\latex{ x }
\latex{ 3 }
\latex{ 2 }
\latex{ 1 }
\latex{ 1 }
\latex{ 3 }
\latex{ y }
Figure 12
Example 2
Let us solve the equation \latex{x\times\log_2x=2} on the set of positive real numbers.
Solution
Let us first rearrange the equation. Since \latex{x \gt 0}, we can divide both sides by \latex{ x }, thus an equation equivalent to the original one is obtained:
\latex{\log_2x=\frac{2}{x}}.
Let us now examine the two functions that are obtained by mapping the set of positive real numbers into the set of real numbers with the expressions on the left-hand-side and on the right-hand-side:
\latex{f:\R^+\rightarrow\R, \;\; f(x)=\log_2x} and \latex{g:\R^+\rightarrow\R,\;\;g(x)=\frac{2}{x}}.
The function \latex{ f } is strictly increasing, the function\latex{ g } is strictly decreasing on the whole domain, thus the values of the functions can be equal at one place at most (Figure 13). It is easy to check that at the place \latex{x = 2} the value of both functions is \latex{ 1 }. It means that the equation has a single solution on the set of positive real numbers and it is \latex{x = 2}.
\latex{y=\frac{2}{x}}
\latex{y=\log_2x}
\latex{ x }
\latex{ 2 }
\latex{ 1 }
\latex{ 1 }
\latex{ y }
Figure 13
Example 3
Let us solve the equation \latex{3^{|x|} + 4^{|x|} = 7^{|x|}} on the set of real numbers.
Solution
Let us divide both sides of the equation by the positive number \latex{4^{|x|}}, and let us use the identities of powers:
\latex{\left(\frac{3}{4}\right)^{\left|x\right|}+1=\left(\frac{7}{4}\right)^{\left|x\right|}}.
Let us examine the two functions, which map from the set of real numbers to the set of real numbers and are defined by the expressions on the right-hand-side and on the left-hand-side, so the functions
\latex{f:\R\rightarrow\R,\;\;f(x)=\left(\frac{3}{4}\right)^{\left|x\right|}+1,\;\;g:\R\rightarrow\R,\;\;g(x)=\left(\frac{7}{4}\right)^{\left|x\right|}}
Their graphs are represented in Figure 14.
On the set of non-negative numbers the function \latex{ f } is strictly decreasing, the function \latex{ g } is strictly increasing, so their values can be equal at one place at most on this set.
At the place \latex{x = 1 f(1) = g(1) =\frac{7}{4} }so \latex{ 1 } is a root of the equation.
Both the function \latex{ f }, and the function \latex{ g } are even, since \latex{f(–x) = f(x)} and \latex{g(–x) = g(x)} for any real x. It implies that the values of the two functions are equal also at the place \latex{x = –1}, and there are no other roots. As if it had another negative root, e.g. \latex{x_1\neq-1} then \latex{0\lt-x_1\neq1} would also be a root, which is not possible.
So the roots of the equation are \latex{ –1 } and \latex{ 1 }.
At the place \latex{x = 1 f(1) = g(1) =\frac{7}{4} }so \latex{ 1 } is a root of the equation.
Both the function \latex{ f }, and the function \latex{ g } are even, since \latex{f(–x) = f(x)} and \latex{g(–x) = g(x)} for any real x. It implies that the values of the two functions are equal also at the place \latex{x = –1}, and there are no other roots. As if it had another negative root, e.g. \latex{x_1\neq-1} then \latex{0\lt-x_1\neq1} would also be a root, which is not possible.
So the roots of the equation are \latex{ –1 } and \latex{ 1 }.
\latex{y=\left(\frac{7}{4}\right)^{\lvert x\rvert}}
\latex{y=\left(\frac{3}{4}\right)^{\lvert x\rvert}+1}
\latex{ 1 }
\latex{ x }
\latex{ -1 }
\latex{ 1 }
\latex{ 2 }
\latex{ y }
Figure 14
Example 4
Let us solve the equation \latex{2^x = 2x} on the set of real numbers.
Solution
Let us again start with defining two functions, which map from the set of real numbers to the set of real numbers:
\latex{f:\R\rightarrow\R,\;\;f(x)=2^x} and \latex{g:\R\rightarrow\R,\;\;g(x)=2x.}
The graphs of the functions are plotted in Figure 15.
Let us use that the function \latex{ f } is convex on the whole set \latex{\R} and the image of the function \latex{ g } is linear. The graphs of the functions \latex{ f } and \latex{ g } have two points in common: \latex{x = 1}, at this place \latex{f(1) = g(1) = 2}, and \latex{x = 2}, at this place \latex{f(2) = g(2) = 4}.
Then because of the convexity there cannot be any more common points, thus our equations has also two roots: \latex{x = 1} and \latex{x = 2}.
Then because of the convexity there cannot be any more common points, thus our equations has also two roots: \latex{x = 1} and \latex{x = 2}.
\latex{y=2^x}
\latex{y=2x}
\latex{ 1 }
\latex{ 2 }
\latex{ x }
\latex{ 2 }
\latex{ 4 }
\latex{ y }
Figure 15
Exercises
{{exercise_number}}. Solve the following equations on the set of real numbers:
- \latex{3^{x}=1-x};
- \latex{2^{-x}=1+x};
- \latex{3^{x}+6^x=9^x};
- \latex{3^{\left|x\right|}+4^{\left|x\right|}=5^{\left|x\right|}};
- \latex{\left(3-2\times\sqrt2\right)^x+\left(3+2\times\sqrt2\right)^x=6^x}.
{{exercise_number}}. Solve the following equations on the set of positive real numbers:
- \latex{x\times\log_3x=8};
- \latex{\log_2x=x-1};
- \latex{3\times\log_{2}{(x+1)}=\log_{\frac{1}{3}}{x}+7};
- \latex{2\times\log_{\frac{1}{3}}x=1-x}.