3.4 Exponential and Logarithmic Equations

The fact that the exponential equation

y = b^x

is equivalent to the logarithmic equation

x = log _b ( y )

may be used to solve some exponential and logarithmic equations.

For example, to solve 2 ^{x - 1} = 8, we can write it in the equivalent logarithmic form

x - 1 = log ₂ ( 8 ) to get the solution x = l + log ₂ ( 8 ) = l + log ₂ ( 2³ ) = 4.

To solve log ₃ x = - 2, we can write it in the equivalent exponential form

x = 3^-2 = 1 / 9.

Exercise 3.4.1

Solve the equations by writing the equation in an equivalent form:

(a) 3^{1 - x} = 9

(b) log ₂ ( x² - x ) = 1

Solution


A one-to-one function f satisies the important property that if a and b are in the domain of f then
f ( a ) = f ( b ) if and only if a = b. Since exponential and logarithmic functions are one-to-one functions, they satisfy this property.

Thus, b^x = b^y if and only if x = y

and

log _b ( x ) = log _b ( y ) if and only if x = y.

Thus, we could solve the equation 5^{1 - x} = 25 by rewriting it as

5^{1 - x} = 5²

and conclude that 1 - x = 2, thus x = - 1.

Likewise, we could solve log ₇ ( x ) = - 2 by replacing the - 2 with log ₇ ( 7^-2 ) to get

log ₇ ( x ) = log ₇ ( 7^-2 ) , concluding that x = 7^-2 = 1 / 49 .

Exercise 3.4.2

Re-solve the problems from Exercise 3.4.1 using the one-to-one property of exponential and logarithmic functions.

Solution

Exercise 3.4.3

Solve e^2x = e^{6 - x}

Solution

Exercise 3.4.4

Solve log₃ ( x² ) - log₃ ( 3 x + 2 ) = 0

Solution

Another way to solve exponential equations is to take a logarithm of each side of the equation using an appropriate base.

For example, to solve 2^x = 7, take log base 2 of each side of the equation to get

log₂ ( 2^x ) = log₂ ( 7 )

Since log₂ ( 2^x ) = x, we get x = log₂ ( 7 ).

This brings up an interesting question: How would one calculate log₂ ( 7 ) ? On a scientific calculator one will find only common logarithm and natural logarithm keys. The answer uses the technique of take the logarithm of both sides using an appropriate base--in this case, base 10 or base e.

For example, to calculate x = log₂ ( 7 ), write the equivalent form 2^x = 7 and take the common logarithm of both sides to get

log ( 2^x )= log ( 7 ) which is equivalent to

x log ( 2 ) = log ( 7 ) . Therefore

x = log ( 7 ) / log ( 2 ). This can be calculated using the log key on a calculator.

This process can be shortened by using the change of base formula for logarithms:

log_a ( x ) = log_b ( x ) / log_b ( a ).

Exercise 3.4.5

Calculate log₅ ( 8 ) to six decimal place accuracy.

Solution

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