x = b^y and y = b^x are inverses. The graphs reflect about the line y = x.

For, x = b^y, y = log_bx, x > 0 and b > 0, b 1

graph	y = bx	x = by or y = logbx
domain	x R	x > 0, x R
range	y > 0, y R	y R
x-intercept	none	1
y-intercept	1	none
horizontal asymptote	x-axis or y = 0	none
vertical asymptote	none	y-axis or x = 0

Note: y = log_bx

Interactive Activity

• Click on the applet to activate.
• Notice the line y = x.
• Move the c and x sliders. Observe the change each slider has on the graph of the exponential function (blue).and logarithmic function (red).

A copy of the above flashlet is provided to study the remainder of the notes for logarithms.

Solving Exponential Functions:

Calculator keystrokes

y = [log] [x-value] [ / ] [log] [b-value] [ENTER]

x = [b-value] [ ^ ] [ y-value ] [ENTER]

b = [x-value] [ ^ ] [ ( ] [ 1 ] [ / ] [y-value] [ ) ] [ENTER]

Interactive Activity

• Change the parameter and "x" sliders. Observe the changes.

General Logarithms

x = b^y y = log_bx

Calculator keystrokes:

log_bx to log₁₀x (log x): [ log ] [ x ] [ ) ] [ ¸ ] [ log ] [ b ] [ ) ]
or
log_bx to log_ex (ln x): [ ln ] [ x ] [ ) ] [ ¸ ] [ ln ] [ b ] [ ) ]

Common Logarithms

x = 10^y y = log₁₀x = log x

Calculator keystrokes:

log x: [ log ] [ x ] [ )]

Natural Logarithms (e = 2.71828...)

x = e^y , y = log_ex = ln x

Calculator keystrokes:

ln x: [ ln ] [ x ] [ ) ]

If b^m = bⁿ, then m = n

Example 1

Solve for x.

Solution

Because the bases are the same, you can use the quotient power property.

Because the bases are the same:

x = -1/2

If log_bM = log_bN, then M = N

Example 1

Solve for y.

log y = log 5x

Solution

y = 5x.

log_bx

For the above, note:

and

log_bx = , since =

log_b(xⁿ) = n log_bx

Example 1: log₂(3²) = 2 log₂ 3

Example 2: log₇(5⁹) = 9 log₇ 5

Example 3: log (5⁹) = 9 log 5

Example 4: ln (5⁹) = 9 ln 5

log_b(bⁿ) = n log_bb = n, n R

log_bbⁿ = n bⁿ = bⁿ or n(log_bb) = n(1) = n

Example 1: log_bb = log_bb¹ = 1 or 1(log_bb) = 1(1) = 1

Example 2: log_bb² = 2 or 2(log_bb) = 2(1) = 2

Example 3: log₁₀10³ = 3 or 3(log₁₀10) = 3(1) = 3

Example 4: log 10⁴ = 4 log 10⁴ = log₁₀10⁴ or 4(log₁₀10) = 4(1) = 4

Example 1:

Solve 2 lnx = 1

Example 2:

Solve e^-x = 5

log_b(xy) = log_bx + log_by

Example 1: log₂(3 · 5) = log₂3 + log₂5

Example 2: log₂(3x) = log_b3 + log_bx

Example 3: log₂(2 · 16) = log₂2 + log₂16 = log₂2¹ + log₂2⁴ = 1 + 4 = 5

log_b(xy)ⁿ = n log_b(xy) = n (log_bx+ log_by)

Example 1: log₂(3 · 5)⁵ = 5 log₂(3 · 5) = 5 (log₂3 + log₂5)

Example 2: log₂(3x)⁵ = 5 log₂(3x) = 5 (log_b3 + log_bx)

Example 3: log₂(2 · 16)⁵ = 5(log₂2 + log₂16) = 5(log₂2¹ + log₂2⁴) = 5 (1 + 4) = 25

log_b(x/y) = log_bx - log_by

Example 1: log₂(3/5) = log₂3 - log₂5

Example 2: log₂(3/x) = log_b3 - log_bx

Example 3: log₂(2/16) = log₂2 - log₂16 = log₂2¹ - log₂2⁴ = 1 - 4 = -3

log_b(x/y)ⁿ = log_bxⁿ - log_byⁿ = n log_bx - n log_by

Example 1: log₂(3/5)⁴ = log₂(3⁴/5⁴) = log₂3⁴ - log₂5⁴ = 4 log₂3 - 4 log₂5

Example 2: log₂(3/x)⁵ = log₂(3⁵/x⁵) = 5 log_b3 - 5 log_bx

Example 3: log₂(2/16)⁵ = log₂(2⁵/16⁵) = 5log₂2 - 5log₂16 = 5log₂2¹ - 5log₂2⁴ = 5(1) - 5(4) = -15

Example 1

Using the laws of logarithms, write in expanded form.

Solution

Example 2

Using the laws of logarithms, write in expanded form.

Solution

Example 3

Evaluate log₃21. Round your answer to 2 decimal places.

Solution

  Let x = log₃21

      3^x = 21

 log 3^x = log 21

x log 3 = log 21

         x = log 21
                log 3

         x2.77

Example 4

Write log₆4 - log₆9 as a single logarithm, then evaluate.

Solution

    log₆4 - log₆9

= log₆(4 · 9)

= log₆36

= log₆6²

= 2

Example 5

Write log₁₀8 - log₁₀1.25 as a single logarithm, then evaluate.

Solution

    log₁₀8 - log₁₀1.25

= log₁₀(8·1.25)

= log₁₀10

= log₁₀10¹

= 1

Example 6

Write log₃108 - log₃4 as a single logarithm, then evaluate.

Solution

Example 7

Write log₃8 - log₃24 as a single logarithm, then evaluate.

Solution

    log₃8 - log₃24

= log₃3^-1

= -1

Example 8

Evaluate.

Solution

Example 9

Evaluate.

Solution

Remember:

Example 10

If log 15 = s, evaluate log 150.

Solution

log 150 = (log 15 · 10)

            = log 15 + log 10

            = s + 1

Example 11

If log 13 = s, evaluate log 1.3.

Solution

            = log 13 - log 10

            = s - 1

Example 14

If log₂x = 9, evaluate.

Solution

Equation Type

One logarithmic expression

Change to exponential form

Example 1

log (x + 1) = 3

Solution

x + 1 = 10³

       x = 1000 - 1

          = 999

Example 2

If log_x2 = -0.25, find x.

Solution: One logarithmic expression - Change to Exponential form.

     log_x2 = -0.25

     x^-0.25 = 2

   (x^-0.25) = 2

(x^-0.25)^-4 = 2^-4

         x¹ = 1/2⁴

           x = 1/16

Example 3

If x = log₈₁3, find x.

Solution: One logarithmic expression - Change to Exponential form.

      x = log₈₁3

  81^x = 3

(3⁴)^x = 3¹

   4x = 1

     x = 1/4

Two logarithmic expressions, constant = 0

Write one equation on each side of the equal sign. Use the equality property to solve.

Example 1

log (3x + 2) - log (2x - 3) = 0

Solution

log (3x + 2) = log (2x - 3)

        3x + 2 = 2x - 3

                x = -5

Three logarithmic expressions, contant 0

Use the product and quotient properties to derive a single logarthimic expression, then use the equality property.

Example 1

log₂(x - 2) + log₂x = log₂3

Solution

log₂(x - 2) + log₂x = log₂3

log₂ x(x - 2) = log₂3

log₂ (x² - 2x) = log₂3

x² - 2x = 3

x² - 2x - 3 = 0

(x - 3)(x + 1) = 0

x = 3 and x = -1

Check                                             Check

log₂(3 - 2) + log₂3 = log₂3                log₂(1 - 2) + log₂1 = log₂3

      log₂(1) + log₂3 = log₂3                    log₂(-1) + log₂1 = log₂3

               log₂(1·3) = log₂3                              log₂(-1·1) = log₂3

                    log₂3 = log₂3                                    log₂-1log₂3

                        LS = RS                                 undefined

                                                                                 LSRS

Therefore, x = 3

Example 2

Express as a function of x and state the domain.

Solution

Recall y = log_cx, x > 0

The domain is x > 0.

Example 3

Solve log x + log (x - 9) = 1.

The trick is to rewrite the two logarithms as a single logarithm and then convert the resulting statement from logarithmic to exponential form. The exponential equation will then be easier to solve.

Caution: Be careful when combining logarithms. You must keep track of any restrictions on the logarithms in the original equation.

Here, log x is defined only if x > 0 and log (x - 9) is defined only if x > 9. Therefore, you must reject any answer less than or equal to 9.

Recall that the restriction on the variable is x > 9. Therefore, the solution is x = 10.

Example 4: Solve for x to 4 decimal places if needed

log₂(x + 5) + log₂(x + 2) = log₂(x + 6)

Solution

First determine the restrictions

x + 5 > 0, x + 2 > 0, x + 6 > 0

x > -5, x > -2, x > -6

If we restrict the domain to x > -2, xR all terms will be valid

log₂(x + 5) + log₂(x + 2) = log₂(x + 6) 

log₂[(x + 5)(x + 2)] = log₂(x + 6) 

(x + 5)(x + 2) = x + 6

x² + 7x + 10 = x + 6

x² + 6x + 4 = 0, x > -2

Factor using the quadratic formula, a = 1, b = 6 and c = 4

x-0.7639 or x-3.2361

Since x > -2, x-0.7639

Two exponential expressions with no common base

Solve algebraically (Example 1) or graphically (Example 2)

Example 1

Solve 2^x+1 = 3^x-1 algebraically. Express your answer as an exact value first; then round your answer to 2 decimal places.

Solution

Now, use your graphing calculator to evaluate this expression correct to 2 decimal places. Note: Use parentheses for the numerator and denominator.

Therefore, x 4.42.

Remember: When you solve exponential equations using logarithms, your solutions are exact values. When you solve exponential equations graphically, your solutions are approximate values!

Example 2

Solve 2^x+1 = 3^x-1 graphically. Round your answer to 2 decimal places.

Solution - This strategy may be used to graph the two sides of any equation with a single variable. The intersect feature of your graphing calculator will determine the solution for the variable.

To solve 2^x+1 = 3^x-1 graphically, you need to determine where the graphs of y = 2^x+1 and y = 3^x-1 cross. The solution of the original equation is the x-coordinate of the intersection point.

Select an appropriate viewing window. Remember, because you are working with exponential functions, the y-values increase quickly. Use the following window settings.

Now, graph y = 2^x+1 and y = 3^x-1.

Use the Intersect feature from the CALCULATE menu to locate the point where the two graphs intersect

Therefore, x 4.42.

Remember: When you solve exponential equations using logarithms, your solutions are exact values. When you solve exponential equations graphically, your solutions are approximate values.