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3.4 Exponential and Logarithmic Equations

Using an equivalence to solve equations

The fact that the exponential equation \(y = b^x\) is equivalent to the logarithmic equation \(x = \log_by\) 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 \(x = 1 + \log_2 ( 8 ) = 1 + \log_2 ( 2^3 ) = 1+3=4\).

To solve \(\log_3 x = -2\), we may write it in the equivalent exponential form \(x = 3^{-2} = \dfrac{1}{9}\).

Exercise 3.4.1

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

  1. \(3^{1-x}=9\)
  2. \(\log_2(x^2-x)=1\)
See Solution

Solutions using the one-to-one property of functions

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 \(\log_bx = \log_by\) if and only if \(x = y\).

Thus, we can solve the equation \(5^{1 - x} = 25\) by rewriting it as \(5^{1 - x} = 5^2\) and conclude that \(1 - x = 2\), thus \(x = -1\).

Likewise, we could solve \(\log_7x = -2\) by replacing the \(-2\) with \(\log_7 ( 7^{-2} )\) to get \(\log_7 x = \log_7 ( 7^{-2} )\) , concluding that \(x = 7^{-2} = \dfrac{1}{49}\) .

Exercise 3.4.2

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

See Solution

Exercise 3.4.3

Solve \(e^{2x} = e^{6 - x}\)

See Solution

Exercise 3.4.4

Solve \(\log_3x^2 - \log_3 ( 3 x + 2 ) = 0\)

See Solution

Taking a logarithm of both sides

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 \(\log_2 2^x = \log_27\) Since \(\log_2 2^x = x\), we get \(x = \log_27\).

Since on a scientific calculator one will find only common logarithm and natural logarithm keys this brings up an interesting question: How would one calculate \(\log_27\)?

The answer is to take the logarithm of both sides using an appropriate base--in this case, base 10 or base e.

For example, to calculate \(x = \log_27\), 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 \(x \log2 = \log7\) . Therefore \(x = \dfrac{\log7}{\log2}\). This can be calculated using the \(\log\) key on a calculator.

One could just as easily use the natural logarithm \(\ln\) instead of the common logarithm \(\log\).

Using a scientific calculator, verify that\(\dfrac{\log7}{\log2} = \dfrac{\ln7}{\ln2}\)

Exercise 3.4.5

Solve \(3^x=11\) to six decimal place accuracy.

See Solution

The change of base formula

The task of finding logarithms to bases other than the common base 10 or the natural base \(e\) is greatly simplified if we use the change of base formula for logarithms.

\[ \log_ax=\frac{\log_bx}{\log_bx} \]

In practice either \(b=10\) or \(b=e\)

Exercise 3.4.6

Use the change of base formula to find \(\log_58\) to six decimal places.

See Solution