Can You Split the Logarithm of a Sum?

A logarithm does not distribute over addition. In general, $\log_b(x+y)$ is not equal to $\log_b(x)+\log_b(y)$. The rule that produces a sum of logarithms applies to a product: $\log_b(xy)=\log_b(x)+\log_b(y)$.

This distinction matters when simplifying expressions, solving equations, or working with a sum written in sigma notation. The safe default is to leave a logarithm of a sum intact unless you can transform the expression using a valid identity.

The logarithm rules

For a real logarithm with base $b$, assume $b>0$ and $b\ne1$. Its input must also be positive. The main product and quotient rules are

$$\begin{aligned} \log_b(xy)&=\log_b(x)+\log_b(y),\\ \log_b\left(\frac{x}{y}\right)&=\log_b(x)-\log_b(y),\\ \log_b(x^r)&=r\log_b(x), \end{aligned}$$

where the inputs to the logarithms are positive. None of these rules says that a sum inside a logarithm can be separated.

The quickest way to remember the product rule is to use the definition of a logarithm. If $p=\log_b(x)$ and $q=\log_b(y)$, then $x=b^p$ and $y=b^q$. Thus $xy=b^{p+q}$, so $\log_b(xy)=p+q$. There is no corresponding exponent rule that turns $x+y$ into an expression involving $p+q$.

A numerical check

Take base 10 and let $x=y=10$. If you incorrectly split the logarithm of their sum, you would get

$$\log_{10}(10+10)=\log_{10}(10)+\log_{10}(10)=2.$$

But the left side is actually

$$\log_{10}(20)\approx 1.3010,$$

not 2. The number 2 is instead the value of $\log_{10}(10\cdot10)$:

$$\log_{10}(100)=2.$$

This illustrates the distinction: adding logarithms corresponds to multiplying their inputs. As another quick check, if $x=y=1$, then $\log_{10}(x+y)=\log_{10}(2)$, while $\log_{10}(x)+\log_{10}(y)=0+0=0$.

What to do with a logarithm of a sum

If the expression is simply $\log_b(x+y)$, first check that $x+y>0$. If the inputs are numbers, add them and then evaluate the logarithm. If they are variables or complicated expressions, there may be no simpler form.

Sometimes factoring is useful. For positive $x$ and $y$, you can factor out $x$ inside the logarithm:

$$\log_b(x+y)=\log_b\left(x\left(1+\frac{y}{x}\right)\right) =\log_b(x)+\log_b\left(1+\frac{y}{x}\right).$$

This is valid because $x+y=x(1+y/x)$ and both logarithm inputs are positive. It is not a rule for splitting addition: the second term still contains a sum. This form can be helpful when one term is much larger than the other, or when you want to factor an expression.

For example, with $x=8$ and $y=2$, base 2 gives

$$\begin{aligned} \log_2(8+2) &=\log_2\left(8\left(1+\frac{2}{8}\right)\right)\\ &=\log_2(8)+\log_2\left(\frac54\right)\\ &=3+\log_2\left(\frac54\right)\\ &\approx 3.3219. \end{aligned}$$

That matches $\log_2(10)\approx3.3219$. It does not equal $\log_2(8)+\log_2(2)=4$, which is $\log_2(8\cdot2)$.

Sums with many terms

The same principle applies to $\log_b(a_1+a_2+\cdots+a_n)$. You cannot turn it into $\sum_{i=1}^n\log_b(a_i)$ in general. For instance, $\log_{10}(1+9)=1$, whereas $\log_{10}(1)+\log_{10}(9)=\log_{10}(9)\approx0.9542$.

By contrast, a logarithm of a product can become a sum:

$$\log_b\left(\prod_{i=1}^n a_i\right)=\sum_{i=1}^n\log_b(a_i),$$

provided every $a_i$ is positive. The product inside the logarithm is essential.

Common mistakes

• Splitting addition: $\log_b(x+y)\ne\log_b(x)+\log_b(y)$ in general. Use the product rule only when the inputs are multiplied.
• Combining the wrong way: $\log_b(x)+\log_b(y)=\log_b(xy)$, not $\log_b(x+y)$.
• Ignoring the domain: For real logarithms, the argument must be positive. Even if one of $x$ or $y$ is negative, $\log_b(x+y)$ may be defined when their sum is positive; however, the separate logarithms $\log_b(x)$ and $\log_b(y)$ may not be.
• Assuming a useful simplification must exist: A logarithm of a sum often has to stay as written. A valid factorization may help, but it does not create a general addition rule.