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3.4 Logarithmic equations

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Revision as of 13:25, 10 September 2008

       Theory          Exercises      

Contents:

  • Logarithmic equations
  • Exponential equations
  • Spurious roots

Learning outcomes:

After this section, you will have learned to:

  • To solve equations that contain powers and by taking logarithms obtain an equation of the first degree.
  • Solve equations that contain logarithm or exponential expressions and which can be reduced to first or second order equations.
  • Deal with spurious roots, and know when they arise.
  • To determine which of two logarithmic expressions is the largest by means of a comparison of bases / arguments.

Basic Equations

Equations where logarithms appear can vary a lot. Here are some examples where the solution is given almost immediately by the definition of a logarithm, that is,

10x=yex=yx=lgyx=lny

(We consider only 10-logarithms or natural logarithms.)

Example 1

Solve the equations

  1. 10x=537 has a solution x=lg537.
  2. 105x=537 gives 5x=lg537, i.e. x=51lg537.
  3. 3ex=5 Multiplication of both sides with ex and division by 5 gives 53=ex, which means that x=ln53.
  4. lgx=3 The definition gives directly x=103=1000.
  5. lg(2x−4)=2 From the definition we have 2x−4=102=100 and it follows that x=52.

Example 2

  1. Solve the equation (10)x=25 .

    Since 10=1012  the left-hand side is equal to (10)x=(1012)x=10x2  and the equation becomes
    10x2=25. 
    This equation has a solution x2=lg25, ie. x=2lg25.
  2. Solve the equation 23ln2x+1=21.

    Multiply both sides by 2 and then subtracting 2 from both sides
    3ln2x=−1.

    Divide both sides by 3

    ln2x=−31.

    Now, the definition directly gives 2x=e−13 , which means that

    x=21e−13=12e13.

In many practical applications of exponential growth or decline there appear equations of the type

ax=b,

where a and b are positive numbers. These equations are best solved by taking the logarithm of both sides

lgax=lgb

and use the law of logarithms for powers

xlga=lgb

which gives the solution  x=lgalgb.

Example 3

  1. Solve the equation 3x=20.

    Take logarithms of both sides
    lg3x=lg20.

    The left-hand side can be written as lg3x=xlg3 giving

    x=lg3lg20(2.727).
  2. Solve the equation  50001.05x=10000.

    Divide both sides by 5000
    1.05x=500010000=2.

    This equation can be solved by taking the lg logarithm of both sides of and rewriting the left-hand side as lg1.05x=xlg1.05,

    x=lg2lg1.05(14.2).

Example 4

  1. Solve the equation  2x3x=5.

    The left-hand side can be rewritten using the laws of exponents giving 2x3x=(23)x and the equation becomes
    6x=5.

    This equation is solved in the usual way by taking logarithms giving

    x=lg6lg5(0.898).
  2. Solve the equation  52x+1=35x.

    Take logarithms of both sides and use the laws of logarithms lgab=blga
    (2x+1)lg52xlg5+lg5=5xlg3,=5xlg3.

    Collect x to one side

    lg5lg5=5xlg3−2xlg5,=x(5lg3−2lg5).

    The solution is

    x=lg55lg3−2lg5.


Some more complicated equations

Equations containing exponential or logarithmic expressions can sometimes be treated as first order or second order equations by considering "lnx" or "ex" as the unknown variable.

Example 5

Solve the equation 6ex3ex+1=5e−x+2.

Multiply both sides by 3ex+1 and e−x+2 to eliminate the denominators

6ex(e−x+2)=5(3ex+1).

Note that since ex and e−x are always positive regardless of the value of x, in this latest step we have multiplied the equation by factors 3ex+1 and e−x+2. Both of these factors are different from zero, so this step cannot introduce new (spurious) roots of the equation.

Simplify both sides of the equation

6+12ex=15ex+5,

where we used e−xex=e−x+x=e0=1. If we treat ex as the unknown variable, the equation is essentially a first order equation which has a solution

ex=31.

Taking logarithms then gives the answer

x=ln31=ln3−1=−1ln3=−ln3.

Example 6

Solve the equation 1lnx+lnx1=1.

The term lnx1 can be written as lnx1=lnx−1=−1lnx=−lnx and then the equation becomes

1lnx−lnx=1,

where we can consider lnx as a new unknown. We multiply both sides by lnx (which is different from zero when x=1) and this gives us a quadratic equation in lnx

1−(lnx)2=lnx,
(lnx)2+lnx−1=0.

Completing the square on the left-hand side

(lnx)2+lnx−1=lnx+212−212−1=lnx+212−45

We continue by taking the root giving

lnx=−2125. 

This means that the equation has two solutions

x=e(−1+5)2ochx=e−(1+5)2. 


Spurious roots

When you solve equations you should also bear in mind that the arguments of logarithms have to be positive and that terms of the type e(...) can only have positive values. The risk is otherwise that you get spurious roots.

Example 7

Solve the equation ln(4x2−2x)=ln(1−2x).

For the equation to be satisfied the arguments 4x2−2x and 1−2x must be equal,

4x2−2x=1−2x ()

and also be positive. We solve the equation () by moving all of the terms to one side

4x2−1=0

and take the root. This gives that

x=−21andx=21.

We now check if both sides of () are positive

  • If x=−21 then both are sides are equal to 4x2−2x=1−2x=1−2−21=1+1=20 .
  • If x=21 then both are sides are equal to 4x2−2x=1−2x=1−221=1−1=00.

So the logarithmic equation has only one solution x=−21.

Example 8

Solve the equation e2x−ex=21.

The first term can be written as e2x=(ex)2. The whole equation is a quadratic with ex as the unknown

(ex)2−ex=21.

The equation can be a little easier to manage if we write t instead of ex,

t2−t=21.

Complete the square for the left-hand side.

t−212−212t−212=21,=43,

which gives solutions

t=21−23andt=21+23. 

Since 31  then 21−2130  and it is only t=21+213  that provides a solution to the original equation because ex is always positive. Taking logarithms finally gives that

x=ln21+23 

as the only solution to the equation.


Exercises

Study advice

The basic and final tests

After you have read the text and worked through the exercises, you should do the basic and final tests to pass this section. You can find the link to the tests in your student lounge.


Keep in mind that:

You may need to spend much time studying logarithms. Logarithms usually are dealt with summarily in high school. Therefore, many college students tend to encounter problems when it comes to calculations with logarithms.

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