The log of a negative number is undefined for all bases (such as log⁡(−3){\displaystyle \log(-3)} or log4⁡(−5){\displaystyle \log {4}(-5)}). Write “no solution. " The log of zero is also undefined for all bases. If you see a term such as ln⁡(0){\displaystyle \ln(0)}, write “no solution. " The log of one in any base (log⁡(1){\displaystyle \log(1)}) always equals zero, since x0=1{\displaystyle x^{0}=1} for all values of x. Replace that logarithm with 1 instead of using the method below. If the two logarithms have different bases, such as log3(x)log4(a){\displaystyle {\frac {log{3}(x)}{log_{4}(a)}}}, and you cannot simplify either one into an integer, the problem is not feasible to solve by hand.

Example 1: Solve the problem log⁡16log⁡2{\displaystyle {\frac {\log {16}}{\log {2}}}}. Start by converting this into one logarithm using the formula above: log⁡16log⁡2=log2⁡(16){\displaystyle {\frac {\log {16}}{\log {2}}}=\log _{2}(16)}. This formula is the “change of base” formula, derived from basic logarithmic properties.

Example 1 (cont. ): Rewrite log2⁡(16){\displaystyle \log _{2}(16)} as 2?=16{\displaystyle 2^{?}=16}. The value of “?” is the answer to the problem. You may need to find it by trial and error:22=2∗2=4{\displaystyle 2^{2}=22=4}23=4∗2=8{\displaystyle 2^{3}=42=8}24=8∗2=16{\displaystyle 2^{4}=8*2=16}16 is what you were looking for, so log2⁡(16){\displaystyle \log _{2}(16)} = 4.

Example 2: What is log3⁡(58)log3⁡(7){\displaystyle {\frac {\log _{3}(58)}{\log _{3}(7)}}}? Convert this into one logarithm: log3⁡(58)log3⁡(7)=log7⁡(58){\displaystyle {\frac {\log _{3}(58)}{\log _{3}(7)}}=\log _{7}(58)}. (Notice that the 3 in each initial log disappears; this is true for any base. ) Rewrite as 7?=58{\displaystyle 7^{?}=58} and test possible values of ?:72=7∗7=49{\displaystyle 7^{2}=77=49}73=49∗7=343{\displaystyle 7^{3}=497=343}Since 58 falls between these two numbers, log7⁡(58){\displaystyle \log _{7}(58)} has no integer answer. Leave your answer as log7⁡(58){\displaystyle \log _{7}(58)}.

For example, start with this problem:“Solve for n if log3⁡(276n)=−6−log3⁡(6){\displaystyle \log _{3}({\frac {27}{6n}})=-6-\log _{3}(6)}. "

If either x or y is negative, there is no solution to the problem. If both x and y are negative, remove the negative signs using the property −x−y=xy{\displaystyle {\frac {-x}{-y}}={\frac {x}{y}}} There are no logarithms of negative numbers in the example problem, so you can continue to the next step.

Use this to expand the left side of the example problem:log3⁡(276n)=log3⁡(27)−log3⁡(6n){\displaystyle \log _{3}({\frac {27}{6n}})=\log _{3}(27)-\log _{3}(6n)} Substitute this back into the original equation:log3⁡(276n)=−6−log3⁡(6){\displaystyle \log _{3}({\frac {27}{6n}})=-6-\log _{3}(6)}→log3⁡(27)−log3⁡(6n)=−6−log3⁡(6){\displaystyle \log _{3}(27)-\log _{3}(6n)=-6-\log _{3}(6)}

The example problem has a new term: log3⁡(27){\displaystyle \log _{3}(27)}. Since 33 = 27, simplify log3⁡(27){\displaystyle \log _{3}(27)} to 3. The full equation is now:3−log3⁡(6n)=−6−log3⁡(6){\displaystyle 3-\log _{3}(6n)=-6-\log _{3}(6)}

3−log3⁡(6n)=−6−log3⁡(6){\displaystyle 3-\log _{3}(6n)=-6-\log _{3}(6)}9−log3⁡(6n)=−log3⁡(6){\displaystyle 9-\log _{3}(6n)=-\log _{3}(6)}log3⁡(6n)=9+log3⁡(6){\displaystyle \log _{3}(6n)=9+\log _{3}(6)}.

In the example problem, the n is still trapped inside the term log3⁡(6n){\displaystyle \log _{3}(6n)}. In order to isolate the n, use the product property of logarithms: loga⁡(bc)=loga⁡(b)+log⁡a(c){\displaystyle \log _{a}(bc)=\log _{a}(b)+\log {a}(c)}log3⁡(6n)=log3⁡(6)+log3⁡(n){\displaystyle \log _{3}(6n)=\log _{3}(6)+\log _{3}(n)} Substitute this back into the full equation:log3⁡(6n)=9+log3⁡(6){\displaystyle \log _{3}(6n)=9+\log _{3}(6)}log3⁡(6)+log3⁡(n)=9+log3⁡(6){\displaystyle \log _{3}(6)+\log _{3}(n)=9+\log _{3}(6)}

log3⁡(6)+log3⁡(n)=9+log3⁡(6){\displaystyle \log _{3}(6)+\log _{3}(n)=9+\log _{3}(6)}log3⁡(n)=9{\displaystyle \log _{3}(n)=9}Since 39 = 19683, n =19683