Why Change the Base?
A logarithm asks for an exponent. For example, \(\log_7(20)\) is the exponent to which \(7\) must be raised to obtain \(20\). The answer is not an integer, and a typical calculator may not have a key for every possible logarithm base.
Most scientific calculators do provide common logarithms, written \(\log(x)\), and natural logarithms, written \(\ln(x)\). A formula connecting these familiar functions to logarithms in other bases lets us evaluate many logarithms directly. It is called the change-of-base formula.
The conditions are essential. A positive base other than \(1\) gives an exponential function that can be inverted to define a logarithm. A logarithm of a nonpositive real argument is not defined as a real number. The bases \(10\) and \(e\), used for common and natural logarithms respectively, satisfy the required base conditions.
Deriving the Change-of-Base Formula
Suppose \(b>0\), \(b\ne 1\), and \(x>0\). Let \(y=\log_b(x)\). By the definition of a logarithm, this means \(b^y=x\). We want to express \(y\) using logarithms that a calculator can evaluate.
For common logarithms, write \(u=\log(b)\). The inverse relationship between the common logarithm and the exponential function with base \(10\) gives \(b=10^u\). Substituting this into \(b^y=x\), and using the exponent identity \((10^u)^y=10^{uy}\), gives \(10^{uy}=x\). Taking the common logarithm of both sides and using that \(\log\) and \(10^x\) are inverse functions gives \(\log(x)=uy\). Since \(u=\log(b)\ne 0\), division by \(u\) yields the formula.
The denominator is not zero: \(\log(b)=0\) would mean \(b=10^0=1\), which the conditions exclude. This also explains why the formula cannot be used with base \(1\).
The same reasoning works with natural logarithms. If \(v=\ln(b)\), then \(b=e^v\). From \(b^y=x\), we obtain \(e^{vy}=x\), so taking \(\ln\) gives \(vy=\ln(x)\). Since \(b\ne1\), \(\ln(b)\ne0\), and therefore:
The formula does not depend on choosing common rather than natural logarithms. The numerator and denominator must use the same kind of logarithm. The conversion works because each ratio equals the same number \(y=\log_b(x)\).
Using Common Logarithms on a Calculator
When a calculator has a \(\log\) key but no key for the required base, enter the argument’s common logarithm divided by the base’s common logarithm. Use parentheses so that the calculator evaluates the entire numerator and denominator before dividing.
Worked Example: Evaluate \(\log_7(20)\)
Here \(b=7\) and \(x=20\). Both satisfy the conditions: \(7>0\), \(7\ne1\), and \(20>0\). Apply change of base using common logarithms:
Using calculator values to ten decimal places, \(\log(20)\approx1.3010299957\) and \(\log(7)\approx0.8450980400\). Divide the numerator by the denominator:
Rounded to six decimal places, \(\log_7(20)\approx1.539502\). Check the result by raising the base to this exponent:
The small difference from exactly \(20\) is due to rounding the logarithm. The check is consistent with the original question: the computed exponent makes \(7\) raised to that power approximately \(20\).
In practice, keep extra calculator digits during the division, then round only the final answer. If the calculator display is rounded before the division, that rounding can affect the quotient. Parentheses also help avoid entering a different expression from the intended ratio.
Worked Example: Evaluate \(\log_3(10)\)
The base \(3\) is positive and not \(1\), and the argument \(10\) is positive, so the logarithm is defined. Use common logarithms:
The calculator gives \(\log(10)=1\) and \(\log(3)\approx0.4771212547\). Thus:
To six decimal places, \(\log_3(10)\approx2.095903\). Check by exponentiation:
The check confirms that the calculated exponent is consistent with the defining equation \(3^y=10\).
Using Natural Logarithms
If a calculator provides an \(\ln\) key, the natural-logarithm version of the formula works just as well. Do not mix \(\ln\) in the numerator with \(\log\) in the denominator: use the same key for both parts of the ratio.
Worked Example: Evaluate \(\log_6(15)\) Using Natural Logarithms
The base \(6\) and argument \(15\) meet the logarithm conditions. Use \(\ln\) for both:
To nine decimal places, \(\ln(15)\approx2.708050201\) and \(\ln(6)\approx1.791759469\). Dividing gives:
Rounded to six decimal places, \(\log_6(15)\approx1.511392\). Check the defining relationship by raising \(6\) to the unrounded quotient:
Thus the exponent found by the calculator reproduces the argument to the displayed precision.
Why the Quotient Works
The derivation can also be viewed as a conversion of units for exponents. The number \(\log_b(x)\) measures the exponent using powers of \(b\). Taking \(\log\) of \(b^y=x\) expresses that same exponent in powers of \(10\): both \(\log(x)\) and \(\log(b)\) are measured in the same common-logarithm scale, so their ratio gives \(y\). Using \(\ln\) changes the scale but not the ratio.
This is why changing the logarithm base in both places leaves the result unchanged. For example, the expressions \(\log(x)/\log(b)\) and \(\ln(x)/\ln(b)\) need not have equal numerators or denominators individually, but each ratio recovers the exponent that satisfies \(b^y=x\).
The calculator result can always be checked against that defining equation. If the calculated value is \(y\), evaluate \(b^y\). A value close to \(x\), with a difference consistent with the number of digits retained, supports the calculation. It is a check rather than a separate proof, since a rounded calculator result is only an approximation.
Common Mistakes and How to Avoid Them
Reversing the ratio. A frequent error is to divide the base logarithm by the argument logarithm. For instance, for \(\log_7(20)\), writing \(\log(7)/\log(20)\) gives approximately \(0.649\), not approximately \(1.540\). That reversed value fails the defining check: \(7^{0.649}\) is approximately \(3.6\), not \(20\). The correct order is always argument logarithm over base logarithm:
Mixing calculator keys. An expression such as \(\log(15)/\ln(6)\) does not follow either version of the formula. Use \(\log(15)/\log(6)\) or \(\ln(15)/\ln(6)\).
Ignoring the conditions. Before entering values, check that the base is positive and not \(1\), and that the argument is positive. For example, \(\log_1(5)\) is not defined because every power of \(1\) equals \(1\), and there is no real exponent \(y\) for which \(1^y=5\).
Rounding too early. If you round both logarithms heavily before dividing, the final quotient may be less accurate than the calculator can provide. Keep the displayed precision through the division and round the final answer to the requested number of decimal places.
A Reliable Calculator Procedure
Confirm that \(b>0\), \(b\ne1\), and \(x>0\).
Use either \(\log\) for both values or \(\ln\) for both values.
Calculate \(\log(x)/\log(b)\) or \(\ln(x)/\ln(b)\), using parentheses if needed.
Retain the calculator’s intermediate digits and round the quotient to the requested precision.
Raise \(b\) to the approximate answer and confirm that the result is close to \(x\).
Why Change of Base Matters
Change of base makes logarithms practical even when a calculator has only a small number of logarithm keys. It also lets different calculations use a consistent scale: a problem can be written in the base that best matches its meaning, while the numerical evaluation uses \(\log\) or \(\ln\).
The main idea is compact: a logarithm is an exponent, and the ratio of two logarithms in the same base recovers that exponent. Remember the domain conditions, keep the argument on top and the base below, and verify the result by exponentiating.
Check Your Understanding
Answer these questions using the definition of a logarithm and the change-of-base formula.
- State the conditions on \(b\) and \(x\) required for \(\log_b(x)\) to be a real logarithm.
- Write \(\log_5(18)\) as a quotient of common logarithms and as a quotient of natural logarithms.
- Why must the numerator and denominator in a change-of-base quotient use the same logarithm key?
- A calculator gives an approximate value \(y\) for \(\log_4(11)\). What calculation checks whether \(y\) is plausible?
- Explain why the expression \(\log(7)/\log(20)\) does not evaluate \(\log_7(20)\).
