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