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 \(x = 3^{-2} = \dfrac{1}{9}\).

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

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

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 \(x = 7^{-2} = \dfrac{1}{49}\) .

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

See SolutionAnother 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\(\dfrac{\log7}{\log2} = \dfrac{\ln7}{\ln2}\)

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.

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

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

See Solution