Logarithmic equations: a common base and original domains

Content updated: October 22, 2026

Only logarithms with the same valid base allow their arguments to be equated directly. Before combining a product or removing logarithms, retain the positivity condition for every argument in the original equation.

Objectives: reduce to a common base, filter algebraic roots, and retain both absolute-value branches.

Prerequisites: the logarithm definition, product and quotient rules, and quadratic equations.

On this page: Idea · Method · Examples · Practice · Further learning

1. Equal logarithms with the same base

For $a>0$, $a\ne1$, and $u(x),v(x)>0$, we have

This follows from injectivity of the logarithm, including bases between 0 and 1. For different bases, logarithms cannot simply be removed. For example, $\log_2 2=\log_3 3=1$, although the arguments 2 and 3 differ.

If the base contains $x$, first require it to be positive and different from 1. The formula applies only at inputs where both sides are defined and have the same base at that input.

2. Combining correctly and filtering roots

State each logarithm's conditions; express constants as logarithms with the same base; combine products or quotients using the known rules; solve the argument equation; and finally check the original domain. A positive product may have two negative factors, but that does not make the original two logarithms defined.

For common logarithms, $\log$ means base 10. In particular, $(1/2)\log(x^2)=\log|x|$ when $x\ne0$, rather than automatically $\log x$. If negative inputs are allowed by the domain, separate the signs of $x$ when recovering the algebraic equation.

3. Worked examples

Example 1. A sum of two logarithms

Solve $\log_3 x+\log_3(x+2)=1$. The conditions $x>0$ and $x+2>0$ are equivalent to $x>0$. On that domain, the equation is equivalent to

Factoring $(x-1)(x+3)=0$ gives $x=1$ or $x=-3$. Accept only 1 because $x>0$. Check: $\log_3 1+\log_3 3=0+1=1$. The number 3 is not a solution here, since $3(3+2)=15$ differs from 3.

Example 2. A logarithm of x²

Solve $\log(x+10)+(1/2)\log(x^2)=2-\log4$. The domain is $x>-10$, $x\ne0$. Using the absolute value and $2=\log100$ yields

For $x>0$, the equation is $x^2+10x-25=0$, with roots $x=-5\pm5\sqrt2$. Accept only $-5+5\sqrt2>0$. For $-10<x<0$, we have $|x|=-x$, so $x^2+10x+25=(x+5)^2=0$, and accept $x=-5$.

Thus the solutions are $x=-5$ and $x=-5+5\sqrt2$. Both satisfy $(x+10)|x|=25$ and belong to the original domain. Replacing log(x²) immediately by 2 log x would lose the negative solution $-5$.

Example 3. A base less than 1

Solve $\log_{1/2}(x+1)=\log_{1/2}(3-x)$. The domain is $-1<x<3$. Equal bases imply $x+1=3-x$, so $x=1$, which is admissible. This is an equality, so no reversal of an inequality is involved.

4. Practice exercises

1. Solve $\log_2(x-1)+\log_2(x+1)=3$.
2. Solve $\log_3(x^2)=2$.
3. Solve $\log_{1/2}(x+2)=\log_{1/2}(4-x)$.

5. Common mistakes

Do not remove logarithms with different bases. Do not replace the domain of a logarithmic sum by that of a logarithm of a product. Retain the absolute value in log(x²), and check algebraic roots before concluding.

6. Related lessons

• Logarithms (English lesson)
• Common and natural logarithms (English lesson)
• The change-of-base formula (English lesson)
• Logarithmic functions (English lesson)
• Logarithmic equations by substitution (English lesson)
• Logarithmic equations using function properties (English lesson)

7. Videos and reference material

• Scribd exercise material 1
• Enrollment link