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Logarithmic equations, simultaneous logarithmic equations and logarithmic inequalities
Now we are going to solve equations and simultaneous equations in which also the logarithm of the unknown appears.
Example 1
Let us solve the following logarithmic equations:
- \latex{\log_{4} (x-2) = 3;}
- \latex{\log_{8} (3x +4) = \log_{8}17;}
- \latex{\log_{7} (x^{2} + x) = \log_{7} (10-2x).}
Solution (a)
The equation has a meaning only if \latex{x-2 \gt 0, \text{i.e.} x \gt 2}.
Based on the definition of logarithm:
Based on the definition of logarithm:
\latex{\begin{align*}x-2 = 4^{3},\\x = 66.\end{align*}}
It corresponds to the interpretation and it can be checked by calculation that it is indeed a solution.
Solution (b)
The equation has a meaning only if (Figure 29)
\latex{\begin{align*}3x+4\gt0,\\x\gt -\frac{4}{3}.\end{align*}}
Since the logarithm function with base \latex{8} is strictly increasing, the logarithms of two expressions are equal if and only if the expressions are equal. In this case
\latex{\begin{align*}3x+4=17,\\3x=13,\\x=\frac{13}{3}.\end{align*}}
The obtained solution is inside the domain, and after substituting \latex{\log_{8} 17} is lies on both sides, so it is indeed a solution.
Figure 29
\latex{ 0 }
\latex{ -\frac{4}{3} }
Solution (c)
The equation has a meaning only if (Figure 30)
\latex{x^2+x\gt0 \;\;\;\text{and}\;\;\;10-2x\gt0,\\x \times(x+1)\gt0,\;\;\;\;\;10\gt2x,\\\underbrace{x\lt-1\;\;\text{or}\;\;x\gt0,\;\;\;\;\;\;\; 5\gt x,}_{x\lt-1 \;\; \text{or}\;\;0\lt x\lt5.}}
Due to the logarithm function with base \latex{7} being strictly increasing
\latex{x^{2} + x = 10 – 2{x},\\x^{2} + 3{x}-10=0,\\x_{1} = 2 \space{\text{and}}\space x_{2} = -5.}
The obtained numbers are elements of the domain. By substituting \latex{x_{1}} we get \latex{\log_{7} 6} on both sides of the equation, and in the case of \latex{x_{2}} we get \latex{\log_{7} 20} on both sides of the equation, so these are indeed solutions.
Figure 30
\latex{ -1 }
\latex{ 1 }
\latex{ 5 }
\latex{ 0 }
\latex{ 1 }
Example 2
Let us solve the following logarithmic equations:
- \latex{\log_{2}x +\log_{2}3=\log_{2}15;}
- \latex{\log (x+13) -\log (x+7)= \log10-\log7;}
- \latex{2\times \log_{3} (x-1) = \log_{3} 4.}
Solution (a)
The equation has a meaning only if \latex{x \gt 0}.
By using identity \latex{ I } let us rewrite the left-hand-side:
By using identity \latex{ I } let us rewrite the left-hand-side:
\latex{\log_{2} (x \times 3) = \log_{2} 15}.
Since the logarithm function is one-to-one:
\latex{3x = 15,\\x=5.}
It also corresponds to the interpretation and by using the identities it can be checked that it is indeed a solution.
Solution (b)
The equation has a meaning if (Figure 31)
\latex{\underbrace{\begin{array}{c}x+13\gt0 \;\;\;\text{and}\;\;\;x+7\gt0,\\x \gt-13;\;\;\;\;\;x\gt-7;\\\end{array}}_{x \gt-7.}}
Based on logarithmic identity \latex{ II } let us rewrite both sides:
\latex{\log\frac{x+13}{x+7}=\log\frac{10}{7}.}
Due to the logarithm function being strictly increasing:
\latex{\frac{x+13}{x+7}=\frac{10}{7}.}
After multiplying across:
\latex{7x + 91 = 10x + 70,\\x=7.}
It corresponds to the interpretation and it can be checked by calculation that it is indeed a solution.
Figure 31
\latex{ -7 }
\latex{ 0 }
\latex{ -13 }
Solution (c)
The equation has a meaning if \latex{x-1 \gt 0, \text{i.e.}\space x \gt 1}. (Figure 32)
By using logarithmic identity \latex{ III } let us rewrite the left-hand-side:
By using logarithmic identity \latex{ III } let us rewrite the left-hand-side:
\latex{\log_{3} (x-1)^2 = \log_{3} 4.}
Due to the strict monotonicity of the logarithm function:
\latex{(x-1)^{2} = 4,\\x^{2}-2x-3 = 0.}
The roots of the quadratic equation are: \latex{x_{1} = 3} and \latex{x_{2} = –1}. The equation has a meaning if \latex{x \gt 1}, therefore \latex{x_{2}} is not a solution.
By substituting \latex{x_{1}} we get \latex{2 \times \log_{3} 2} on the left-hand-side, and since \latex{4=2^{2}}, therefore \latex{x_{1}} is a solution.
By substituting \latex{x_{1}} we get \latex{2 \times \log_{3} 2} on the left-hand-side, and since \latex{4=2^{2}}, therefore \latex{x_{1}} is a solution.
Figure 32
\latex{ 1 }
\latex{ 0 }
Example 3
Let us solve the following logarithmic equations:
- \latex{\log (10x-2) -2\times\log (x+1)= \log2;}
- \latex{\log \sqrt{x-2} -\log (x-5)+ \log2=0;}
- \latex{\log_{3} (x-1) -\log_{3}(6x-5)+\log_{3} (2-3x)=2.}
Solution (a)
The equation has a meaning if (Figure 33)
\latex{\underbrace{\begin{array}{c}10x-2\gt0 \;\;\;\text{and}\;\;\;x+1\gt0,\\x \gt\frac{1}{5};\;\;\;\;\;x\gt-1;\\\end{array}}_{\text{Both are satisfied if }x \gt\frac{1}{5}.}}
Let us rewrite the left-hand-side of the equation with the help of the logarithmic identities:
\latex{\log\frac{10x-2}{(x+1)^{2}}=\log2.}
Due to the logarithm function being strictly increasing:
\latex{\frac{10x-2}{(x+1)^{2}}=2.}
After multiplying across:
\latex{\begin{align*}10x – 2 = 2 \times(x + 1)^{2},\\10x-2=2x^{2} + 4x + 2,\\0 = 2x^{2}-6x + 4,\\0 = x^{2}-3x + 2.\end{align*}}
The roots are: \latex{x_{1} = 1} and \latex{x_{2} = 2}. Both of the roots correspond to the interpretation, and checking also supports that these are solutions.
Figure 33
\latex{ 0 }
\latex{ \frac {1}{5} }
\latex{ -1 }
Solution (b)
The equation has a meaning if (Figure 34)
\latex{\underbrace{\begin{array}{c}x-2\gt0 \;\;\;\text{and}\;\;\;x-5\gt0,\\x \gt2 ;\;\;\;\;\;x\gt5;\\\end{array}}_{x\space\gt \space5.}}
Let us rewrite the left-hand-side of the equation by using the logarithmic identities, and on the right-hand-side let us use that \latex{\log 1 = 0}:
\latex{\log\frac{2\times \sqrt{x-2}}{x-5}=\log1.}
Due to the logarithm function being strictly increasing:
\latex{\frac{2\times \sqrt{x-2}}{x-5}=\log1.}
After multiplying across let us square:
\latex{\begin{align*}2\times\sqrt{x-2}=x-5,\\4\times(x-2)=x^{2}-10x+ 25,\\4x-8=x^{2}-10x+25,\\ 0=x^{2}-14x+33.\end{align*}}
The solutions of the quadratic equation are: \latex{x_{1} = 3} and \latex{x_{2} = 11}.
Only the second root corresponds to the interpretation, checking implies that it is indeed a solution.
Only the second root corresponds to the interpretation, checking implies that it is indeed a solution.
Figure 34
\latex{ 5 }
\latex{ 2 }
\latex{ 0 }
Solution (c)
The equation has a meaning if (Figure 35)
\latex{{\begin{array}{c}x-1\gt0 \;\;\;\text{and}\;\;\;6x-5\gt0,\;\;\;\text{and}\;\;\;2-3x\gt0,\\x \gt1 ;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;x\gt\frac{5}{6};\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\frac{2}{3}\gt x.\\\end{array}}}
There is no real number for which the equation can be defined, so there is no solution.
Figure 35
\latex{ 0 }
\latex{ \frac {2}{3} }
\latex{ \frac {5}{6} }
\latex{ 1 }
Example 4
Let us solve the following simultaneous equations on the set of real numbers:
- \latex{\begin{rcases}\begin{align*}5\times\log{x}+3\times\log{y=2}\\2\times\log{x}-\log{y=3}\end{align*}\end{rcases}};
- \latex{\begin{rcases}\begin{align*}\log{(x+1)}+\log{(y-3)=1}\\\log{(y-1)}-\log{x=0}\end{align*}\end{rcases}}.
Solution (a)
The simultaneous equations have a meaning if \latex{x \gt 0, y \gt 0}.
The example becomes easy to overview if we introduce new variables: let \latex{a = \log x, b = \log y}. Then the simultaneous equations are:
\latex{\begin{rcases}\begin{align*}{5a+3b=2}\\{2a-b=3}\end{align*}\end{rcases}}.
By adding \latex{3} times the second one to the first one:
\latex{11a = 11}.
It implies \latex{a = 1, b = –1}. By substitution:
\latex{{\begin{array}{c}\log x=1\;\;\;\text{and}\;\;\;\log y=-1,\;\;\;\\x =10 ,\;\;\;\;\;\;\;y=\frac{1}{10}.\;\;\;\;\;\\\end{array}}}
By checking we can confirm that the number pair \latex{\left(10;\frac{1}{10}\right) } is indeed a solution.
\latex{\begin{array}{r}5a+3b=2\\6a-3b=9\\\hline11a=11\end{array}}
Solution (b)
The simultaneous equations have a meaning only if \latex{x\gt0\space \text{and}\space y\gt3}.
Let us rewrite the first equation by using the logarithmic identities:
Let us rewrite the first equation by using the logarithmic identities:
\latex{\log\left[(x+1)\times(y-3)\right]=\log10}.
Due to the strict monotonicity of the logarithm function:
\latex{(x + 1)\times(y-3) = 10}. \latex{\left(1\right)}
Let us rearrange the second equation:
\latex{\log (y-1) = \log x}.
Due to the strict monotonicity of the logarithm function:
\latex{y-1 = x}.
By substituting \latex{x} into equation \latex{(1)}:
\latex{(y-1 + 1)\times(y - 3) = 10}.
By expanding the parentheses and rearranging:
\latex{y^{2}-3y-10 = 0}.
The roots of the quadratic equation are: \latex{y_{1} = 5, y_{2} = –2}.
Due to the interpretation \latex{y_{2}} is not a solution. \latex{x = 4} belongs to \latex{y_{1}}.
It can be checked by calculation that the number pair \latex{(4; 5)} is indeed a solution.
Due to the interpretation \latex{y_{2}} is not a solution. \latex{x = 4} belongs to \latex{y_{1}}.
It can be checked by calculation that the number pair \latex{(4; 5)} is indeed a solution.
Example 5
Let us solve the following inequalities on the set of real numbers:
- \latex{\log_{5}(5x-3)\gt \log_{5}3=\log_{2}(2x+3);}
- \latex{\log_\frac{2}{3} (2x-1)\gt\log_\frac{2}{3} x;}
- \latex{\log_{12} x+\log_{12}(x+1)\lt1.}
Solution (a)
The inequality has a meaning if \latex{x\gt\frac{3}{5}}.
Since the logarithm function with base \latex{5} (Figure 36) is strictly increasing, the inequality holds if
\latex{5x-3\gt 2x+ 3,\\x\gt 2.}
By comparing it to the interpretation the solution is:
\latex{x\gt 2.}
Figure 36
\latex{y= \log_{5} x}
\latex{ 1 }
\latex{ x }
\latex{ 1 }
\latex{ y }
\latex{\log_{5}b=\log_{5}c; \space b,c\gt0\\\Updownarrow\\b\gt c}
Solution (b)
The inequality has a meaning if \latex{x\gt\frac{1}{2}}.
Since the function \latex{f(x)=\log_\frac{2}{3}x} (Figure 37) is strictly decreasing, the inequality holds if:
\latex{2x-1\lt x,\\x\lt 1.}
By comparing it to the interpretation the solution is:
\latex{\frac{1}{2}\lt x\lt 1.}
Figure 37
\latex{y= \log_\frac{2}{3} x}
\latex{ 1 }
\latex{ x }
\latex{ 1 }
\latex{ y }
\latex{\log_{\frac{2}{3}}b\gt\log_{\frac{2}{3}}c; \space b,c\gt0\\\Updownarrow\\b\lt c}
Solution (c)
The inequality has a meaning if \latex{x \gt 0}.
Let us apply the logarithmic identities on the left-hand-side, and let us rewrite the right-hand-side:
\latex{\log_{12}[x\times(x+1)]\lt\log_{12}12}.
The logarithm function with base \latex{12} is strictly increasing (Figure 38), the inequality holds if
Figure 38
\latex{y= \log_{12} x}
\latex{ 1 }
\latex{ x }
\latex{ 1 }
\latex{ y }
\latex{x\times(x+1)\lt{12},\\x^{2}+x-12\lt0.}
The roots of the quadratic equation \latex{x^{2} + x-12 = 0} are: \latex{x_{1} = 3} and \latex{x_{2} =-4}.
The solution of the quadratic inequality is:
The solution of the quadratic inequality is:
\latex{–4 \lt x \lt 3}.
By comparing it to the interpretation the solution of the inequality is (Figure 39):
\latex{0 \lt x \lt 3}.
Figure 39
\latex{y=x^{2}+x-12}
\latex{ -4 }
\latex{ -2 }
\latex{0 }
\latex{2 }
\latex{3 }
x
y
\latex{ 2 }
Example 6
Let us solve the following equations on the set of real numbers:
- \latex{\log_{2}x+\log_{8}x-\log_{16}(4x)=6;}
- \latex{\log_{4} [\log_{3}(\log_{2}x)]=0;}
- \latex{x^{\log{x-3}}=0.01.}
Solution (a)
The equation has a meaning if \latex{x \gt 0}. Let us convert to the logarithm with base \latex{2}:
\latex{\log_{2}x+\frac{\log_{2}x}{\log_{2}8}-\frac{\log_{2}(4x)}{\log_{2}16}=6.}
The denominators can be calculated, and the third numerator can be rewritten based on logarithmic identity I:
\latex{\log_{2}x+\frac{\log_{2}x}{3}-\frac{\log_{2}4+\log_{2}x}{4}=6.}
When multiplying by \latex{12}:
\latex{\begin{align*}12\times\log_{2}x+4\times\log_{2}x-3\times(2+\log_{2}x)=72,\\13\times\log_{2}x=78,\\ \log_{2}x=6,\\ x=2^{6},\\x=64.\end{align*}}
By substitution we can confirm that \latex{64} is indeed a solution of the equation.
Solution (b)
The equation has a meaning if (Figure 40)
\latex{\begin{array}{rcl}x\gt0 \;\;\text{and}\;\;\log_2x\gt0\;\;\text{and}\;\;\log_3{(\log_2 x)}\gt0,\\x\gt1;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\log_2x\gt1,\\x\gt2;\\\underbrace{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}_{x\gt2.}\end{array}}
Based on the definition of logarithm, if
\latex{\log_{4} [\log_{3}(\log_{2}x)]=0,\space\text{then}\\\log_{3}(\log_{2}x)=1.}
Let us apply the definition again:
\latex{\log_{2}x=3,\\x=2^{3},\\x=8.}
It also corresponds to the interpretation and by substituting – calculating the logarithms one by one – it turns out that \latex{x = 8} is indeed a solution of the equation.
Figure 40
\latex{ 1 }
\latex{ 2 }
\latex{ 0 }
Solution (c)
The equation has a meaning if \latex{x \gt 0}. Since the logarithm function is strictly increasing, let us take the logarithm of both sides with base \latex{10}:
\latex{\log x^{\log x-3} = \log 0.01.}
The left-hand-side can be rewritten based on the logarithmic identity III, and the right-hand-side can be calculated:
\latex{(\log x-3)\times \log x = –2,\\(\log x)^2-3\times\log x+2=0.}
Let us write a new variable instead of \latex{\log x, \text{let} \log x = y}. The new form of the equation is:
\latex{y^{2} – 3y + 2 = 0}.
The solutions of the quadratic equation are: \latex{y_{1} = 2} and \latex{y_{2} = 1}.
By substitution:
By substitution:
\latex{\log x=2 \quad \text{or}\quad \log x=1,}
\latex{x=100;\qquad x=10.}
\latex{x=100;\qquad x=10.}
We can check both numbers, they are indeed solutions of the equation.
Exercises
{{exercise_number}}. Solve the following equations on the set of real numbers:
- \latex{2=\log_{3}(x+1)};
- \latex{\log_\frac{1}{2}(3-x)=-2};
- \latex{\log (x-2) + \log (x + 3) = \log 24};
- \latex{\log_{3} (x-4) + \log_{3} (x-2) = 1};
- \latex{\log (x-3) + \log (x + 4) = \log (5x + 9)};
- \latex{\log_{5} (3x-1) – \log_{5} (x + 2) = 1};
- \latex{\log_{3} (2x + 10) = 2\times \log_{3} (x + 1)};
- \latex{2\times\log 2 + (1 + x)\times \log 3-\log (3^{2x} + 27) = 0;}
- \latex{\log\left[x^{\log \left(x^{\log x}\right) }\right]=\log x };
- \latex{\sqrt[3]{x^{-3+\log_{2}2x^{2}}}=\frac{x^{2}}{4}}.
{{exercise_number}}. Solve the following simultaneous equations on the set of real numbers:
- \latex{\begin{rcases}\begin{align*}\log{x}+\log{y=2}\\\log{y}-\log{x=\log 25}\end{align*}\end{rcases}};
- \latex{\begin{rcases}\begin{align*}5\times\log_{2}x-3\times\log_{3}{y=9}\\2\times\log_{2}{x}+\log_{3}{y=8}\end{align*}\end{rcases}};
- \latex{\begin{rcases}\begin{align*}\log_{3}x+\log_{3}y=2+\log_{3}2\\ \log_{12}\frac{x}{y}=1\end{align*}\end{rcases}};
- \latex{\begin{rcases}\begin{align*}x^{\log_{8}y}+ y^{\log_{8}x}=4\\\log_{4}x-\log_{4}y=1\end{align*}\end{rcases}};
- \latex{\begin{rcases}\begin{align*}10^{1+\log(x+y)}=50\\\log(x-y)+\log(x+y)=2-\log 5\end{align*}\end{rcases}};
- \latex{\begin{rcases}\begin{align*}\log_{5}x+3^{\log_3y}=2\\x^{y}=\frac{1}{125}\end{align*}\end{rcases}};
{{exercise_number}}. Solve the following inequalities on the set of real numbers:
- \latex{\log_{17}(3x+4)\lt 1};
- \latex{3\gt\log_\frac{1}{2}(2x+1)};
- \latex{\log_\frac{3}{4}\frac{5-x}{3x+1}\gt 0};
- \latex{\log(4 x^{2}+4x-2)\gt 0};
- \latex{\log\left(2^{x-1}-28 \right)\gt 2 };
- \latex{-2\gt\log_\frac{1}{3}\frac{x-3}{4x-1}};
- \latex{\log_\frac{1}{2}\frac{\mid x-4\mid}{2x+3}\gt3};
- \latex{1\gt \left(\frac{1}{2}\right)^{\log_{2-x}\left(x^{2}-4x + 3\right)}};
- \latex{\log_{x} (x^{2} + x-4) \lt 1}.
Puzzle
Give the value of the product: \latex{(x-\log_{a} a^{a})\times(x-\log_{b} b^{b})\times(x-\log_{c} c^{c})\times…\times(x-\log_{z} z^{z} )}.