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Exponential equations,simultaneous exponential equations and exponential inequalities

Example 1

Solve the following equations on the set of real numbers:

\latex{4^{x} = 8\times \sqrt[3]{2}}
\latex{5^{x}\times 5= 25^{x-2}}

Solution (a)

Let us express both sides as a power of \latex{ 2 }, and let us perform the operations:

\latex{(2^{2})^{x}= 2^{3} \times 2^{\frac{1}{3} }} , \latex{2^{2x}= 2^{\frac{10}{3} }} .

Since the exponential function is strictly increasing, the two powers can be equal to each other if and only if the exponents (indices) are equal too:

\latex{2\times = \frac{10}{3}} , \latex{x= \frac{5}{3}}.

We performed equivalent steps, so the solution is: \latex{x= \frac{5}{3}}.

Solution (b)

Let us act similarly as before, now let us express both sides as a power of \latex{ 5 }:

\latex{5^{x+1} = (5^{2})^{x-2}} ,
\latex{5^{x+1} = 5^{2x-4}}.

Due to the exponential function being strictly increasing:

\latex{x+ 1= 2\times - 4} ,
\latex{\times = 5} .

With quick mental arithmetic it can be checked that the result is indeed a solution.

Example 2

Let us solve the equation \latex{7^{x^{2}- 2\times - 8 } = 1} on the set of real numbers.

Solution (b)

Since \latex{ 1 } is the power of every positive number with an exponent of \latex{ 0 }, the equation can be written as follows:

\latex{7^{x^{2}- 2\times - 8 } = 7^{0}} .

Due to the exponential function being strictly increasing:

\latex{x^{2} - 2\times - 8= 0} ,
\latex{x_{1,2} = \frac{2\pm \sqrt{4+ 32} }{2}} ,
\latex{x_{1} = 4} , \latex{x_{2} = - 2} .

With calculation it can be checked that both obtained values are roots of the original equation.

Example 3

Let us solve the following equation on the set of real numbers:

\latex{5^{x} + 5^{x+1} + 5^{x+2} = 31} .

Solution

There is a sum on the left-hand-side of the equation; it cannot be expressedas a power. Let us try to rearrange the exponents:

\latex{5^{x} + 5^{x} \times 5^{1} + 5^{x} \times 5^{2} = 31} .

By factoring out the common factor, then by combining in the parentheses:
\latex{5^{x} \times (1+5^{-1}+5^{2})= 31},
\latex{5^{x}\times 31= 31} ,
\latex{5^{x} = 1} ,
\latex{x= 0} .
The solution can be checked with calculation.

Example 4

Let us solve the following equation on the set of real numbers:

\latex{9^{x+1} = 4\times 6^{x}} .

Solution

Again we cannot express both sides of the equation as a power of the same number. Every base is a multiple of \latex{ 2 } or \latex{ 3 }, therefore let us rewrite both sides as powers of \latex{ 2 } and \latex{ 3 }:
\latex{(3^{2})^{x+1}= 2^{2} \times (2\times 3)^{x}} ,
\latex{3^{x+2}= 2^{2} \times 2^{x} \times 3^{x}} .

Let us divide both sides by the non-\latex{ 0 } (definitely positive) \latex{3^{x}} :
\latex{3^{x}= 2^{x+2}} .
Let us again divide both sides, now by \latex{2^{x+2}} :
\latex{\frac{3^{x+2} }{2^{x+2} } = 1} , \latex{(\frac{3}{2} )^{x+2}= 1} ,

\latex{(\frac{3}{2} )^{x+2}= (\frac{3}{2} )^{0}} , \latex{x+2=0}

\latex{x=-2}

The validity of the solution can be checked by calculation.

Example 5

Let us solve the following equation on the set of real numbers:

\latex{4^{x} - 7\times 2^{x} - 8= 0} .

Solution

We can try to express each term as a power of \latex{ 2 } to no avail; we will not reach our goal. Let us realise that there is a close relationship between the two terms containing unknowns:
\latex{4^{x} = (2^{2})^{x}= 2^{2x} = (2^{x})^{2}} .
If we introduce a new variable, the equation becomes simple.

Let \latex{2^{x} = y} . Thus the equation is:

\latex{y^{2} - 7y- 8= 0} ,
\latex{y_{1,2} = \frac{7\pm \sqrt{49+32} }{2}} ,
\latex{y_{1} =8 ; y_{2} = 1} .

\latex{y_{2}} is not a solution, because \latex{y\gt 0} . By substitution: if \latex{2^{x} = 8, x=3} .
The solution of the equation is \latex{x=3}, which can be checked by a quick calculation.

Example 6

Let us solve the following simultaneous equations on the set of real numbers:

\latex{\begin{rcases}5\times 3^{x}-2\times 2^{y}=7 \\ 2\times 3^{x}+2^{y}=10\end{rcases};}

Solution

The example can be solved easily if we introduce new variables.
Let \latex{3^{x} = a} and \latex{2^{y} = b , a\gt 0 , b\gt 0} . With this substitution we have to solve the simultaneous equations

\latex{\begin{rcases}5a-2b=7 \\ 2a+b=10\end{rcases};}

Let us express \latex{ b } from the second equation:

\latex{b= 10-2a}.

Let us substitute it into the first one:

\latex{5a- 2\times (10-2a)= 7} ,
\latex{5a-20+4a=7} ,
\latex{9a=27}.

Hence \latex{a=3} and \latex{b=4}. By substitution: \latex{3^{x}=3} and \latex{2^{y}=4, x=1} and \latex{y=2}. It can be checked by calculation that the number pair \latex{(1;2)} is indeed a solution.

Example 7

Let us solve the following simultaneous equations on the set of real numbers:

\latex{\begin{rcases}4\times 2^{2x}=4 \\ 10^{xy}=10^{10}\times 100^{x+1}\end{rcases};}

Solution

There are the powers of \latex{ 2 } in the first equation, and the powers of \latex{ 10 } in the second equation, therefore let us rearrange the equations separately.

From the first one:

\latex{2^{2} \times 2^{2x} = 2^{2y}} ,
\latex{2^{2+2x} = 2^{2y}}.

Due to the exponential function being strictly increasing:

\latex{2+2x=2y} ,
\latex{y=x+1}.

From the second one:

\latex{10^{xy} = 10^{10} \times 10^{2x+2}} ,
\latex{10^{xy} = 10^{2x+12}}.

Due to the exponential function being strictly increasing:

\latex{xy=2x+12}.

Let us substitute y obtained from the first one:

\latex{x\times (x+1)= 2x+12}.

By rearranging the quadratic equation:

\latex{x^{2} -x-12=0} ,
\latex{x_{1,2} = \frac{1\pm \sqrt{1+48} }{2}}
\latex{x_{1} =4; x_{2}=-3}.

Hence \latex{y_{1} =5} and \latex{y_{2} =-2}, so the solutions are the number pairs \latex{(4;5)} and \latex{(-3;-2)}.

Example 8

Let us solve the following inequalities on the set of real numbers:

\latex{3^{2x-1} \gt 9;}

\latex{4^{|x|+3} \lt 16^{x}}

Solution (a)

Since \latex{9=3^{2}} and the exponential function with base \latex{ 3 } is strictly increasing. (Figure 13)
\latex{3^{2x-1} \gt 3^{2}} if and only if \latex{2x-1\gt 2 , x\gt \frac{3}{2} }.

Thus the solution of the inequality is:

\latex{x\gt \frac{3}{2}}.

Solution (b)

Let us also rewrite the right-hand-side as a power of \latex{ 4 }:

\latex{4^{|x|+3} \lt 4^{2x} } .

Since the exponential function with base \latex{ 4 } is strictly increasing:

\latex{|x|+3\lt 2x}.

Because of the absolute value we distinguish two cases:

Case I: If \latex{x\geq 0} , then
\latex{x+3\lt 2x}
\latex{3\lt x}

Figure 14/a

Case II: If \latex{x\lt 0}
\latex{-x+3\lt 2x}
\latex{3\lt 3x}
\latex{1\lt x}

Figure 14/b

\latex{ 0 }

\latex{ 3 }

\latex{ 0 }

\latex{ 1 }

Since no solution resulted from case II, the result is: \latex{x\gt 3}.

Example 9

Let us solve the following inequalities on the set of real numbers:

\latex{(\frac{1}{2} )^{x+3}\leq 4;}

\latex{(\frac{3}{5} )^{x^{2}-3x-10 }\gt 1}.

Solution (a)

The number \latex{ 4 } on the right-hand-side of the inequality can be expressedas a power of \latex{\frac{1}{2}:}

\latex{(\frac{1}{2} )^{x+3} \leq (\frac{1}{2} )^{-2}}.

Since the exponential function with base \latex{\frac{1}{2}} is strictly decreasing,the inequality holds exactly if \latex{x+3\geq -2}, from which the solution is: \latex{x\geq -5}.

Solution (b)

The number \latex{ 1 } on the right-hand-side of the inequality can be written as:

\latex{1= (\frac{3}{5} )^{0}}.

\latex{(\frac{3}{5} )^{x^{2}-3x-10 }\gt (\frac{3}{5} )^{0}} ,

the exponential function with base \latex{\frac{3}{5}} is strictly decreasing (Figure 15) therefore:

\latex{x^{2} -3x-10\lt 0} .

The roots of the quadratic equation \latex{x^{2} -3x-10=0} are \latex{x_{1} =5} and \latex{x_{2} =-2} , thus the solution of the inequality is (Figure 16) :
\latex{-2\lt x\lt 5}.

\latex{y=(\frac{3}{5} )^{x}}

\latex{ 1 }

\latex{ x }

\latex{ 1 }

\latex{ y }

Figure 15

\latex{y=x^{2}-3x-10}

\latex{ 2 }

\latex{ 5 }

\latex{-2 }

\latex{ y }

\latex{ x }

Figure 16

Exercises

{{exercise_number}}. Solve the following equations on the set of real numbers:

\latex{4^{2x-1} \times 2^{x} = 16^{x};}

\latex{9^{x-1}=81\times \sqrt{3};}

\latex{10^{x^{2}-4x+3 } = 1};

\latex{6^{2x^{2} } \times 6^{7x} = 6^{15};}

\latex{27\times 2^{x} = 8\times 3^{x};}

\latex{125\times 3^{x-1} = 3\times 5^{x+1};}

\latex{2^{x-1}+2^{x+1}=20};

\latex{3^{x}+3^{x+2} +3^{x-1} =\frac{31}{3};}

\latex{25^{x}+5= 6\times 5^{x};}

\latex{9^{x} + 6\times 3^{x}-27=0};

\latex{2^{x+4} +2^{x+3}+2^{x} =5^{x+1}-5^{x};}

\latex{(\frac{1}{4} )^{3x}- (\frac{1}{8} )^{x-1}= 128};

\latex{9^{x} + \sqrt{x^{2}+2 } -4\times 3^{x-1}\\ +\sqrt{x^{2} +2} =69};

\latex{4^{\sin^{2}x } +4^{\cos^{2} x} =4};

\latex{x^{2x^{2}-7x+6} =1}.

{{exercise_number}}. Solve the following simultaneous equations on the set of real numbers:

\latex{\begin{rcases}3 \times 5^{x} - 2^{y} = 7 \\ 2 \times 5^{x} + 3 \times 2^{y} = 34\end{rcases};}

\latex{\begin{rcases}4\times 7^{x}+2^{y}=20\\ -3\times 7^{x}+4\times 2^{y}=61\end{rcases};}

\latex{\begin{rcases}5^{2x}\times 5^{y}=5^{5}\\ 6^{x-y}=6\end{rcases};}

\latex{\begin{rcases}(5^{x})^{y}-2^{y}=\frac{1}{25} \\ 27^{x}=\frac{3}{3^{y}}\end{rcases};}

\latex{\begin{rcases}(2^{x^{2}+2x-4})^{y2}=\frac{1}{16} \\ \sqrt{3^{x}} =\sqrt[6]{27^{y-1}}\end{rcases};}

\latex{\begin{rcases}16^{2x}+16^{2y}=36 \\ 16^{x+y}=8\times \sqrt{2}\end{rcases};}

\latex{3^{x}+3^{y-\frac{1}{2} }=4 \\ (2x-y)^{2}=\frac{9}{4};}

\latex{\begin{rcases}x^{y^{2}-2y-35}=1 \\ 2x+y=4\end{rcases};}

\latex{\begin{rcases}2^{2x-2y}+2^{x-y}=2 \\ 2^{2x+1}+(\frac{1}{2} )^ {2y-1}=5\end{rcases};}

{{exercise_number}}. Solve the following inequalities on the set of real numbers:

\latex{(\frac{8}{7} )^{5x-4}\lt 1;}

\latex{(\frac{7}{9} )^{2x-3}\geq \frac{9}{7};}

\latex{9\times \sqrt[4]{3}\leq (\frac{1}{3} )^{x};}

\latex{5^{x} \leq 125^{|x|-1} ;}

\latex{(\frac{3}{4} )^{x^{2}-12 }\geq (\frac{3}{4} )^{x};}

\latex{5^{\frac{x+1}{x-2} } \lt 25;}

\latex{9\times 2^{x+4} \leq 2\times 3^{x+5} ;}

\latex{25^{x}- 5^{x} \gt 5^{x+1} -5;}

\latex{(x^{2}-x+1 )^{x-2}\gt 1}.

Puzzle

Solve the following equation on the set of real number pairs: \latex{2\times \cos ^{2} \frac{x^{2}+3y }{6} = 3^{x}+3^{-x}}.

Mathematics 11.

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