# Solving Logarithmic Equations

Equations involving logarithms and unknown variables can often be solved by employing the definition of the logarithm, as well as several of its basic properties:

\( \log_x(a) + \log_x(b) = \log_x(ab) \)

\(\log_x(a) - \log_x(b) = \log_x\big(\frac ab\big) \)

\(a\log_x(b) = \log_x(b^a) \)

\(\log_x(a) = \frac{\log_y(a)}{\log_y(x)} = \frac1{\log_a(x)}\) for any positive real number \(y \)

\(x^{\log_x(a)} = a\).

Common mistakes to watch out for include

\(\log_x(a) \cdot \log_x(b) \neq \log_x(ab) \)

\(\frac{\log_x(a)}{\log_x(b)} \neq \log_x\big(\frac{a}{b}\big)\).

The general strategy is to consolidate the logarithms using these properties, and then to take both sides of the equation to the appropriate power in order to eliminate the logarithms if possible.

## Solving Logarithmic Equations - Basic

For many equations with logarithms, solving them is simply a matter of using the definition of \( \log x \) to eliminate logarithms from the equation and convert it into a polynomial or exponential equation.

Find \( x \) if \( \log_2(3x+1) = 4 \).

By the definition of the logarithm,

\[\begin{align} 3x+1 &= 2^4 \\&= 16 \\ 3x &= 15 \\ x &= 5.\ _\square \end{align}\]

Another way to view this solution is that we took \( 2 \) to the power of the left side and got \( 3x+1 \), and took \( 2 \) to the power of the right side and got \( 2^4 = 16 \). More complicated logarithmic equations are often simplified by exponentiating both sides.

Other equations can be simplified using other properties of logarithms. One difficulty that arises is that eliminating logarithms and solving the resulting equation can introduce spurious solutions. These solutions violate the principle that the argument of the log function must always be positive. It is generally wise to check solutions by plugging them into the original equation and making sure that both sides are defined.

Find all \( x\) such that \( \log_3(x-12) +\log_3(x-6) = 3. \)

Simplify the given equation as follows:

\[\begin{align} \log_3(x-12) + \log_3(x-6) &=3 \\ \log_3\big((x-12)(x-6)\big) &= 3\\ (x-12)(x-6) &= 27 \\ x^2-18x-45 &=0\\ (x-15)(x-3) &= 0. \end{align}\]

This produces two potential solutions \( x=3, x=15\). But note that \( x = 3 \) is not an actual solution, as \( \log_3(3-12) \) is undefined. The only actual solution is \( x=15 \). \(_\square\)

The step in the solution above that was not reversible was the first one: although \( (x-12)(x-6)\) is positive if \( x =3\), \( x-12 \) and \( x-6\) are not themselves positive.

## Solving Logarithmic Equations - Intermediate

More complicated logarithmic equations often involve more than one base. It can help to introduce unknowns to solve for the logarithms first. Another useful identity is \( \log_x(y) = \frac{\log_z(y)}{\log_z(x)} \), especially since \( z\) can be chosen to be whatever simplifies the problem.

Suppose \( a,b\) are positive real numbers such that

\[\log_a(10)+\log_b(100) = \log_{ab}(1000000).\]

Show that \( a = b \) or \( a^2=b \).

Rewrite this as

\[\frac{\log(10)}{\log(a)} + \frac{\log(100)}{\log(b)} = \frac{\log(1000000)}{\log(ab)},\]

where the logs are all to the base \( 10 \). This simplifies to

\[\frac1{\log(a)}+\frac2{\log(b)} = \frac6{\log(ab)}.\]

Let \( m = \log(a) \) and \( n = \log(b) \). Then \(\log(ab) = m+n\), so

\[\begin{align} \frac1{m}+\frac2{n} &= \frac6{m+n} \\ n+2m &= \frac{6mn}{m+n} \\ (2m+n)(m+n) &= 6mn \\ 2m^2+3mn+n^2 &= 6mn \\ 2m^2-3mn+n^2 &= 0 \\ (2m-n)(m-n) &= 0. \end{align}\]

So \( m=n\) or \( 2m=n\). In the first case, \( \log(a)=\log(b) \), so \( a=b\). In the second case, \( 2\log(a) = \log(b) \), so \( \log(a^2)=\log(b) \), so \(a^2=b \). \(_\square \)

Equations involving exponents can often be simplified by taking logarithms:

Solve for \( x \) if

\[\left( \frac{x}2 \right)^{\log_2(x)} = 8x.\]

Take \(\log_2\) of both sides:

\[\begin{align} \log_2\left(\left(\frac{x}2 \right)^{\log_2(x)}\right) &= \log_2(8x) \\ \log_2(x)\log_2\left(\frac x2\right) &= \log_2(x)+3 \\ \log_2(x)\big(\log_2(x)-1\big) &= \log_2(x)+3, \end{align}\]

and substituting \( y = \log_2(x) \) turns this into \( y(y-1)=y+3\), or \( y^2-2y-3 = 0 \). So \( y = 3 \) or \( y = -1 \), which leads to \( x = 8 \) or \( x = \frac 12 \). \(_\square\)

Logarithms can also make computation easier in certain practical situations. For example, logarithms with base \(10\) give information about the number of decimal digits in a number.

Given that

\[\begin{align} \log_{10}(2) &= 0.3010299\ldots \\ \log_{10}(3) &= 0.4771212\ldots \\ \log_{10}(7) &= 0.8450980\ldots, \end{align}\]

which is bigger, \( 12^{50} \) or \( 7^{64} \)? How many decimal digits do these two numbers have?

Compute

\[\begin{align} \log_{10}\big(2^{100}3^{50}\big) &= 100\log_{10}(2)+50\log_{10}(3) \\ &= 30.10299\ldots + 23.85656\ldots \\ &= 53.9595\ldots \\\\ \log_{10}\big(7^{64}\big) &= 64\log_{10}(7) \\ &= 64(0.8450980\ldots) \\ &= 54.0862\ldots. \end{align}\]

So \( 7^{64} \) is bigger. The number of decimal digits of \( x \) is \( \lfloor \log_{10}(x) \rfloor +1 \), so \( 2^{100}3^{50} \) has \( 54 \) digits and \( 7^{64} \) has \( 55 \) digits. \(_\square\)

Note that this is much easier than multiplying the numbers and comparing them directly. For applications that require large numbers (such as RSA encryption), this is very important.

**Cite as:**Solving Logarithmic Equations.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/solving-logarithmic-equations/