Lesson Solving logarithmic equations
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<H2>Solving logarithmic equations</H2> Logarithmic equations contain the unknown variable under the logarithmic function. The simplest logarithmic equation has the form {{{log(a,f(x))}}} = {{{log(a,g(x))}}}, where {{{a}}} is a real positive number not equal to 1. To solve the equation {{{log(a,f(x))}}} = {{{log(a,g(x))}}}, you need 1) to solve the equation {{{f(x)}}} = {{{g(x)}}}; 2) to check the found roots for conditions {{{f(x)>0}}}, {{{g(x)>0}}}. Only those roots that satisfy these conditions are the solutions of the original equations. Other roots are not the solutions of the original equations. There are two major methods for solving logarithmic equations: 1) A method of transforming the original equation to the form {{{log(a,f(x)) = log(a,g(x))}}} and then replacing it by the equation {{{f(x) = g(x)}}}, and 2) A method of introducing a new variable. Problems below show how to use these methods. <H3><U>Problem 1</U></H3>Solve the equation {{{log(5,(x^2-7x+11))}}} = {{{log(5,(5-2x))}}}. <B>Solution</B> First, pass from the given equation to the equation {{{x^2-7x+11 = 5-2x}}}. Then, simplify this equation by moving the terms from the right side to the left side with the opposite signs and collecting the common terms. As a result, you get the quadratic equation {{{x^2 - 5x + 6 = 0}}}. You can solve this equation by applying the quadratic formula (see the lesson <A HREF = http://www.algebra.com/algebra/homework/quadratic/lessons/Introduction-Into-Quadratics.lesson>Introduction into Quadratic Equations</A>) {{{x}}} = {{{(5+-sqrt(5^2-4*1*6))/2}}} = {{{(5+-sqrt(25-24))/2}}} = {{{(5+-sqrt(1))/2}}}. The two roots are {{{x[1] = 3}}} and {{{x[2] = 2}}}. Now check these two roots for conditions {{{x^2-7x+11 > 0}}} and {{{5-2x > 0}}}. The value x=3 does not satisfy these conditions, while the value x=2 does. <B>Answer</B>. x=2. <B><U>Exercise 1</U></B> Solve yourself the equation {{{log(3,(x^2-3x-5))}}} = {{{log(3,(7-2x))}}}. <H3><U>Problem 2</U></H3>Solve the equation {{{log(10,(x+6)) + log(10,(2x+3))}}} = {{{log(10,(7+2x))}}}. <B>Solution</B> Using logarithm properties (sum of logarithms is equal to the logarithm of the product; see for example the lesson <A HREF = http://www.algebra.com/algebra/homework/logarithm/Properties-of-the-logarithm.lesson>Properties of the logarithm</A>), you can transform the original equation into the form {{{log(10,((x+6)*(2x+3)))}}} = {{{log(10,(7+2x))}}}, Then consider the equation {{{(x+6)*(2x+3)}}} = {{{7+2x}}}. Simplify this equation step by step. Open brackets in the left side, move the terms from the right side to the left side with the opposite signs and collect the common terms: {{{2x^2+12x+3x+18}}} = {{{7+2x}}}, {{{2x^2+13x+11}}} = {{{0}}}. Eventually, you get the quadratic equation, which you can solve using quadratic formula (see the lesson <A HREF = http://www.algebra.com/algebra/homework/quadratic/lessons/Introduction-Into-Quadratics.lesson>Introduction into Quadratic Equations</A>) {{{x}}} = {{{(-13+-sqrt(13^2-4*2*11))/(2*2)}}} = {{{(-13+-sqrt(169-88))/4}}} = {{{(-13+-sqrt(81))/4}}}. Two roots are {{{x=-5.5}}} and {{{x=-1}}}. Now you should check these two roots for conditions {{{x+6>0}}}, {{{2x+3>0}}} and {{{7+2x>0}}}. The value x=-1 does satisfy these conditions, while the value x=-5.5 doesn't. <B>Answer</B>. The solution is x=-1. <B><U>Exercise 2</U></B> Solve yourself the equation {{{log(10,(x+4)) + log(10,(2x+3))}}} = {{{log(10,(1-2x))}}}. <H3><U>Problem 3</U></H3>Solve the equation {{{(log(2,x))^2 + 2*log(2,x) - 3}}} = {{{0}}}. <B>Solution</B> Let us introduce a new variable {{{y =log(2,x)}}}. Then you get the quadratic equation {{{y^2+2*y-3 = 0}}}. It has the roots {{{y[1]}}} = {{{-3}}}, {{{y[2]}}} = {{{1}}}. Now the problem is reduced to the solution the equations {{{log(2,x)}}} = {{{-3}}} and {{{log(2,x)}}} = {{{1}}}. The first of these two equations has the root x = {{{1/8}}}. The second equation has the root x = 2. <B>Answer</B>. The equation has two solutions: x = {{{1/8}}} and x = 2. <B><U>Exercise 3</U></B> Solve yourself the equation {{{(log(2,x))^2 + log(2,x) + 1}}} = {{{7/log(2,(0.5x))}}} introducing new variable. <H3><U>Problem 4</U></H3>If xy = 64 and {{{ log (x, y) + log (y, x) = 5/2 }}}, find x and y. <B>Solution</B> <pre> {{{log (x, y) + log (y, x)}}} = {{{5/2 }}}. (1) Well known rule for the base change of logarithms says that {{{log (y, x)}}} = {{{1/log(x,y)}}}. Therefore, the equation (1) becomes {{{log (x, y)}}} + {{{1/log (x, y)}}} = {{{5/2 }}}. (2) To solve it, introduce new variable z = {{{log (x, y)}}}. Then the equation (2) becomes {{{z + 1/z}}} = {{{5/2}}}, or {{{z^2 - (5/2)z + 1}}} = 0, or {{{2z^2 - 5z + 2}}} = 0. Use the quadratic formula to find its roots. The roots are {{{z[1]}}} = 2 and/or {{{z[2]}}} = {{{1/2}}}. 1. {{{z[1]}}} = 2 ====> {{{log(x,y)}}} = 2 ====> y = x^2. Substitute it into the equation xy = 64, and you will get {{{x^3}}} = 64, which implies x = 4. In this case, the solution of the original system is x = 4, y = 16. 2. {{{z[2]}}} = {{{1/2}}} ====> {{{log(x,y)}}} = {{{1/2}}} ====> y = x^(1/2). Substitute it into the equation xy = 64, and you will get x^(3/2) = 64, which implies {{{x^3}}} = {{{64^2}}} and hence x = 16. In this case, the solution of the original system is x = 16, y = 4. </pre> <B>Answer</B>. There are two solutions: 1) x=4, y= 16, and 2) x=16, y = 4. <H3><U>Problem 5</U></H3>Solve an equation log x(2x-1)=1. <B>Solution</B> <pre> log x(2x-1) = 1 (by default, the base of the logarithm assumes to be 10) ====> x*(2x-1) = 10 ====> 2x^2 - x - 10 = 0 ====> {{{x[1,2]}}} = {{{(1 +- sqrt(1^2 + 4*2*10))/(2*2)}}} = {{{(1 +- sqrt(81))/4}}} = {{{(1 +- 9)/4}}}. {{{x[1]}}} = {{{(1+9)/4}}} = {{{10/4}}} = 2.5. It works: log (2.5*(2*2.5-1)) = log (2.5*4) = log (10) = 1. {{{x[2]}}} = {{{(1-9)/4}}} = {{{-8/4}}} = -2. It works too, since log ((-2)*(2*(-2)-1) = log ((-2)*(-5)) = log(10) = 1. </pre> <B>Answer</B>. The given equation has two roots x= 2.5 and x= -2. <H3>Problem 6</H3>Solve an equation {{{8^(x-2)}}} = {{{2/25}}}. <B>Solution</B> <pre> Your starting equation is {{{8^(x-2)}}} = {{{2/25}}}. It is equivalent to {{{2^(3*(x-2))}}} = {{{8/100}}}. Take logarithm base 10 of both sides 3*(x-2)*log(2) = 3*log(2) - 2 Simplify (3x - 6)*log(2) = 3*log(2) - 2 (3x - 6 - 3)*log(2) = -2 (3x - 9)*log(2) = -2 3x - 9 = - {{{2/log((2))}}} 3x = - {{{2/log((2))}}} + 9 x = - {{{2/(3*log((2)))}}} + 3 = 0.785381 (approximately; rounded). <U>ANSWER</U> <U>CHECK</U>. The left side of the original equation is {{{8^(0.785381-2)}}} = 0.080000 = {{{8/100}}} = {{{2/25}}}. ! Correct ! </pre> <H3>Problem 7</H3>Solve an equation {{{4*log(16,(x))}}} = {{{log(4,(x+12))}}}. <B>Solution</B> <pre> The original equation is {{{4*log(16,(x))}}} = {{{log(4,(x+12))}}}. Note that the domain, where both sides are defined, is the set of positive real x: { x | x > 0 }. Due to properties of logarithms, the original equation is the same as {{{4*(1/2)*log(4,(x))}}} = {{{log(4,(x+12))}}}, or, equivalently, {{{2*log(4,(x))}}} = {{{log(4,(x+12))}}}, which, in turn, is equivalent to {{{log(4,(x^2))}}} = {{{log(4,(x+12))}}}. It implies x^2 = x + 12 x^2 - x - 12 = 0, (x-4)*(x+3) = 0. Of the two roots, x= 4 and x= -3, only positive x= 4 is in the domain and is, therefore, the solution. <U>ANSWER</U>. x= 4. </pre> <H3>Problem 8</H3>Solve an equation {{{10^(4*log(x))}}} - {{{7*(10^(2*log(x)))}}} + {{{10}}} = {{{0}}}. <B>Solution</B> <pre> First, since "x" is under the logarithm function, it must be positive: x > 0. So, the domain for this equation is the set {x | x > 0 }. Second, {{{10^log(x)}}} = x; therefore, the original equation in the domain x > 0 is equivalent to this simple equation {{{x^4 - 7x^2 + 10}}} = 0. Factor left side: (x^2-5)*(x^2-2) = 0. Formally, it has 4 solutions {{{sqrt(5)}}}, {{{-sqrt(5)}}}, {{{sqrt(2)}}} and {{{-sqrt(2)}}}. Of these four numbers, only positive numbers belong to the domain, so the final <U>answer</U> is two numbers {{{sqrt(5)}}} and {{{sqrt(2)}}}. </pre> My other lessons in this site on logarithms, logarithmic equations and relevant word problems are - <A HREF=http://www.algebra.com/algebra/homework/logarithm/what-is-the-logarithm.lesson>WHAT IS the logarithm</A> - <A HREF=http://www.algebra.com/algebra/homework/logarithm/Properties-of-the-logarithm.lesson>Properties of the logarithm</A> - <A HREF=http://www.algebra.com/algebra/homework/logarithm/change-of-base-formula-for-logarithms.lesson>Change of Base Formula for logarithms</A> - <A HREF=https://www.algebra.com/algebra/homework/logarithm/Evaluate-logarithms-without-using-a-calculator.lesson>Evaluate logarithms without using a calculator</A> - <A HREF=https://www.algebra.com/algebra/homework/logarithm/Simplifying-expressions-with-logarithms.lesson>Simplifying expressions with logarithms</A> - <A HREF=https://www.algebra.com/algebra/homework/logarithm/Solving-advanced-logarithmi%D1%81-equations.lesson>Solving advanced logarithmic equations</A> - <A HREF=https://www.algebra.com/algebra/homework/logarithm/Solving-really-interesting-and-educative-problem-on-logarithmic-equation-containing-a-HUGE-underwater-stone.lesson>Solving really interesting and educative problem on logarithmic equation containing a HUGE underwater stone</A> - <A HREF=https://www.algebra.com/algebra/homework/logarithm/Proving-equalities-with-logarithms.lesson>Proving equalities with logarithms</A> - <A HREF=https://www.algebra.com/algebra/homework/logarithm/Solving-logarithmic-inequalities.lesson>Solving logarithmic inequalities</A> - <A HREF=http://www.algebra.com/algebra/homework/logarithm/Using-logarithms-to-solve-real-world-problems.lesson>Using logarithms to solve real world problems</A> - <A HREF=https://www.algebra.com/algebra/homework/logarithm/Solving-problem-on-Newton-Law-of-cooling.lesson>Solving problem on Newton Law of cooling</A> - <A HREF=https://www.algebra.com/algebra/homework/logarithm/Population-growth-problems.lesson>Population growth problems</A> - <A HREF=https://www.algebra.com/algebra/homework/logarithm/Radioactive-decay-problems.lesson>Radioactive decay problems</A> - <A HREF=https://www.algebra.com/algebra/homework/logarithm/Carbon-dating-problems.lesson>Carbon dating problems</A> - <A HREF=https://www.algebra.com/algebra/homework/logarithm/Bacteria-growth-problems.lesson>Bacteria growth problems</A> - <A HREF=https://www.algebra.com/algebra/homework/logarithm/A-medication-decay-in-a-human%27s-body.lesson>A medication decay in a human's body</A> - <A HREF=https://www.algebra.com/algebra/homework/logarithm/Problems-on-appreciated-depreciated-values.lesson>Problems on appreciated/depreciated values</A> - <A HREF=https://www.algebra.com/algebra/homework/logarithm/Inflation-and-Salary-problems.lesson>Inflation and Salary problems</A> - <A HREF=https://www.algebra.com/algebra/homework/logarithm/Miscellaneous-problems-on-exponential-growth-decay.lesson>Miscellaneous problems on exponential growth/decay</A> - <A HREF=https://www.algebra.com/algebra/homework/logarithm/Problems-on-discretely-compound-accounts.lesson>Problems on discretely compound accounts</A> - <A HREF=https://www.algebra.com/algebra/homework/logarithm/Problems-on-continuously-compound-accounts.lesson>Problems on continuously compound accounts</A> - <A HREF=https://www.algebra.com/algebra/homework/logarithm/Tricky-problem-on-solving-a-logarithmic-system-of-equations.lesson>Tricky problem on solving a logarithmic system of equations</A> - <A HREF=https://www.algebra.com/algebra/homework/logarithm/Uninterrupted-withdrawing-money-from-a-retirement-fund.lesson>Entertainment problem: Uninterrupted withdrawing money from a retirement fund</A> - <A HREF=https://www.algebra.com/algebra/homework/logarithm/Entertainment-problems-on-logarithms.lesson>Entertainment problems on logarithms</A> - <A HREF=https://www.algebra.com/algebra/homework/logarithm/Entertainment-problems-on-exponential-growth.lesson>Entertainment problems on exponential growth</A> - <A HREF=https://www.algebra.com/algebra/homework/logarithm/Upper-level-problem-on-solving-logarithmic-equations.lesson>Upper level problems on solving logarithmic equations</A> - <A HREF=http://www.algebra.com/algebra/homework/logarithm/OVERVIEW-of-lessons-on-logarithms-logarithmic-eqns-and-relevant-word-probs.lesson>OVERVIEW of lessons on logarithms, logarithmic equations and relevant word problems</A> Use this file/link <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson>ALGEBRA-I - YOUR ONLINE TEXTBOOK</A> to navigate over all topics and lessons of the online textbook ALGEBRA-I.
