How to Solve Logarithmic Equations: Definition and Properties

Step-by-step Guide to Solve Logarithmic Equations

Here is a step-by-step guide to solving logarithmic equations:

Step 1: Understand the Basics

• Logarithmic Function: A logarithm is the inverse of an exponential function. It tells us the power to which a base number must be raised to obtain another number. The logarithm of a number \(b\) to the base \(a\) is denoted as \(log_a​\left(b\right)\) and is equivalent to the statement \(a^{log_a​\left(b\right)}=b\).

For instance, \(log _2\left(8\right)=3\) because \(2^3=8\).

Step 2: Familiarize Yourself with the Properties of Logarithms

Before attempting to solve logarithmic equations, it’s essential to be well-versed in logarithm properties:

• Product Rule: \(log _b\left(m×n\right)\)\(=\)\(log _b\left(m\right)\)\(+\)\(log _b\left(n\right)\)
• Quotient Rule: \(log _b\left(\frac{m}{n}\right)\)\(=\)\(log _b\left(m\right)\)\(-\)\(log _b\left(n\right)\)
• Power Rule: \(log _b\left(m^{n}\right)\)\(=n \ ×\) \(log _b\left(m\right)\)
• Change of Base Rule: \(log_b​\left(a\right)\) \(=\)\(\frac{log_c​\left(a\right)}{log_c​\left(b\right)}\) for any positive value of \(c\), where \(c≠1\).

Step 3: Rewrite in Exponential Form

If you are given a logarithmic equation and you’re not sure how to proceed, first try expressing it in its equivalent exponential form. This transformation often simplifies the equation or provides a clearer perspective.

Example: For the equation \(log_2​​\left(x\right)=3\), in exponential form, it becomes \(2^3=x\), which implies \(x=8\).

Step 4: Use Properties to Simplify

When faced with multiple logarithms in an equation, use the properties of logarithms to combine or simplify them.

Example: To solve \(log(x)+log(3x)=2\), using the product rule, it becomes \(log(3x^2)=2\).

Step 5: Solve for the Variable

Once the equation is simplified, either convert the equation into an exponential form to solve directly or use other algebraic methods.

Continuing our example: From \(log(3x^2)=2\) with the base \(10\) implied, We get \(3x^2=10^2\), which is \(3x^2=100\). Solve this quadratic equation to get values of \(x\).

Step 6: Check the Domain

Remember that you can only take the logarithm of a positive number. Always check the domain to ensure the solutions are valid.

Using our example, if you find a negative value for \(x\), it would be extraneous since the logarithm of a negative number is undefined.

Step 7: Verify Your Solution

Plug the obtained solution back into the original logarithmic equation to ensure it holds true.

Step 8: Understand the Graphical Implications

Understanding the graph of logarithmic functions can provide visual insights:

• The graph of a basic logarithmic function \(y=\)\(log_b​\left(x\right)\) is a curve that increases slowly, passing through \((1,0)\).
• Logarithmic functions will always have a vertical asymptote at \(x=0\) since the logarithm of \(0\) is undefined.

You’ll be better equipped to tackle and solve logarithmic equations by understanding these steps and properties. As with all mathematical techniques, practice is key. Engage in various problems to hone your skills and deepen your comprehension.

Examples:

Example 1:

Evaluate: \(log_2​\left(8\right)\)

Solution:

The logarithmic expression asks the following question: “To what power must \(2\) be raised to get \(8\)?”

Since \(2^3=8\), the value for which this is true is \(3\).

Thus, \(log_2​\left(8\right)\)\(=3\).

Example 2:

Simplify the expression: \(log _5\left({25}\right)\)\(-\)\(log _5\left(5\right)\)

Solution:

To simplify this, we’ll uses the properties of logarithms, specifically the Quotient Rule which states:

\(log _b\left(\frac{m}{n}\right)\)\(=\)\(log _b\left(m\right)\)\(-\)\(log _b\left(n\right)\)

Applying the Quotient Rule to our expression: \(log _5\left({25}\right)\)\(-\)\(log _5\left(5\right)\)\(=log _b\left(\frac{25}{5}\right)\)

This simplifies to: \(log _5\left(5\right)\)

Given that the logarithm asks, “To what power must the base (in this case, \(5\)) be raised to obtain the number?”, and since \(5^1=5\), the answer is \(1\).

Thus, the simplified expression is: \(log _5\left(5\right)=1\)

A Couple More Worked Examples

Example 1: Solve \(\log_3(x) + \log_3(x – 2) = 1\).

Start by using the product rule to combine the left side: \(\log_3\big(x(x-2)\big) = 1\). Rewrite in exponential form: \(x(x-2) = 3^1 = 3\). Expand: \(x^2 – 2x – 3 = 0\). Factor: \((x – 3)(x + 1) = 0\). So \(x = 3\) or \(x = -1\). Now check the domain, both \(x\) and \(x – 2\) must be positive. If \(x = -1\), then \(\log_3(-1)\) is undefined. So the only valid answer is \(x = 3\).

Example 2: Solve \(\log_2(x + 5) – \log_2(x – 1) = 2\).

Use the quotient rule: \(\log_2\left(\frac{x+5}{x-1}\right) = 2\). Rewrite in exponential form: \(\frac{x+5}{x-1} = 2^2 = 4\). Multiply both sides by \(x – 1\): \(x + 5 = 4(x – 1)\). Solve: \(x + 5 = 4x – 4\), so \(3x = 9\), so \(x = 3\). Check the domain, \(x + 5 = 8 > 0\) and \(x – 1 = 2 > 0\). So \(x = 3\) is valid.

Common Mistakes (and How to Avoid Them)

Forgetting the domain check. This is the single most common mistake on log equations. Always, always, check that every expression inside a logarithm is positive after you solve.

Misusing the product rule. \(\log(a) + \log(b) = \log(ab)\), not \(\log(a + b)\). The plus sign on the outside becomes a multiplication on the inside. Many students get this backwards under pressure.

Skipping the verification step. Even when your algebra is correct, you can land on an extraneous solution that doesn’t actually satisfy the original equation. Plug your answer back in. It takes 30 seconds and saves a lot of wrong answers.

Recommended Reading

Logarithmic equations are typically introduced near the end of Algebra II, and they show up regularly on the SAT, ACT, CLEP College Algebra, and any precalculus test. If you want a slower, more thorough walk through logs (and the rest of Algebra II), Algebra II for Beginners is the resource I’d recommend. It explains the underlying ideas with the same patience I tried to bring above, and it gives you plenty of practice problems to lock the skills in.

Frequently Asked Questions About Logarithmic Equations

What is an extraneous solution in logarithmic equations?

An extraneous solution is a value that comes out of your algebra but does not work in the original equation, usually because it makes a log argument zero or negative. For log(x) + log(x − 3) = 1, combining gives x(x − 3) = 10, so x = 5 or x = −2. Only x = 5 works, since log(−2) is undefined.

How do you solve logarithmic equations step by step?

First combine the logs on each side into a single log using the product, quotient and power rules. Next, set the arguments equal (if both sides are single logs with the same base) or rewrite the equation in exponential form. Solve the algebra, then check each answer keeps every log argument positive. For log₂(x + 1) = 3, x + 1 = 2³, so x = 7.

How do you solve logarithmic equations with logs on both sides?

When each side is a single log with the same base, set the arguments equal. For log₃(2x + 1) = log₃(x + 6), write 2x + 1 = x + 6, so x = 5. Check it: both arguments equal 11, which is positive, so the answer stands. If a side has more than one log, combine them first with the product or quotient rule.

How do you solve logarithmic equations with different bases?

Rewrite the logs in one base, either with the change-of-base formula, logₐ(x) = log(x)/log(a), or by spotting that one base is a power of the other. For log₂(x) = log₄(16) + 1, note that 16 = 4², so log₄(16) = 2. The equation becomes log₂(x) = 3, which gives x = 2³ = 8.

How do you solve logarithmic equations with e?

Logs with base e are natural logs, written ln, and you undo them by making both sides a power of e. For ln(x − 2) = 3, rewrite it as x − 2 = e³, so x = e³ + 2, which is about 22.09. The same domain rule applies: the argument x − 2 must be positive, and here it is.

How do you solve logarithmic equations with a number in front?

Use the power rule to move the number into the exponent: n·log(x) = log(xⁿ). For 2 log₅(x) = log₅(36), rewrite the left side as log₅(x²), so x² = 36. That gives x = 6 or x = −6, but −6 is extraneous because the original log₅(x) needs a positive input, so the only solution is x = 6.