Exponential equations can be solved with logarithms. Likewise, logarithmic equations can be solved with exponents, as well as with the identities present related to both types of functions. The combination of these techniques are very useful in problem-solving in this area. Many types of problems will be shown, in order to showcase the diversity of the problems found when studying equations with logarithms.
As with other content related to exponents and logarithms, it is important to remember:
Logarithms and exponents are inverse operations. Therefore, a logarithm and exponent with the same base will undo each other.
This section will focus on solving equations where logarithms have products or quotients, or must be converted to having products and quotients in the logarithms' arguments.
Example 1a: Solve the equation for \(x\): \(\log_{8}(4) + \log_{8}(2x) = \log_{8}(15x^3) - \log_{8}(5x)\).
Solution: The two logarithms on both sides of the equation can be combined using identities:
$$\log_{8}(8x) = \log_{8}(3x^2)$$
Take the \(8\)th exponent of both sides of the equation to eliminate the logarithms:
$$8x = 3x^2$$
\(x = 0\) appears to be a solution, but in the original equation, we get \(0\) as the argument for three of the four logarithms, so this is an extraneous solution, or a false solution. If we factor out the \(x\)s common to both sides, we get
$$8 = 3x \Rightarrow$$ $$x = \frac{8}{3}$$
This solution, on the other hand, causes no problems in the original equation.
Example 1b: Another solution to the problem in Example 1a will be shown here. Note that this is a more subtle solution than the first.
Break up the second logarithm on the left side to get
$$\log_{8}(4) + \log_{8}(2) + \log_{8}(x) = \log_{8}(15x^3) - \log_{8}(5x)$$
Combine the first two logarithms by multiplying the arguments:
$$\log_{8}(8) + \log_{8}(x) = \log_{8}(15x^3) - \log_{8}(5x) \Rightarrow$$ $$1 + \log_{8}(x) = \log_{8}(15x^3) - \log_{8}(5x)$$
Combine the two logarithms on the right-hand side in the same way as was done in Example 1a:
$$1 + \log_{8}(x) = \log_{8}(3x^2)$$
The remaining two logarithms can be combined, because they have the same base:
$$1 = \log_{8}(3x^2) - \log_{8}(x) \Rightarrow$$ $$1 = \log_{8}(3x)$$
Remove the logarithm by the same method that was used in Example 1a:
$$8 = 3x \Rightarrow$$ $$x = \frac{8}{3}$$
That was a rather complicated example that required many steps, but most logarithmic equations require the combination of multiple identities related to logarithms and exponents, rather than just one identity.
Example 2: Solve this equation for \(x\): \(\log_{x}(4) + \log_{x}(7) = \log_{x}(2) + 2\).
Solution: Keep in mind the variable is the base of all the logarthms but does not appear elsewhere. Combine the two logarithms on the left-hand side as follow:
$$\log_{x}(4) + \log_{x}(7) = \log_{x}(4 \cdot 7) = \log_{x} = 28$$]
This can be done by means of a logarithm identity, giving
$$\log_{x}(28) = \log_{x}(2) + 2$$
Move the logarithm on the right-hand side to the left-hand side and combine the two logarithms by creating a quotient:
$$\log_{x}(28) - \log_{x}(2) = 2 \Rightarrow$$ $$\log_{x}(14) = 2$$
Take the \(x\)th exponent of both sides:
$$14 = x^2 \Rightarrow$$ $$x = \pm \sqrt{14}$$
The solution \(-\sqrt{14}\) is extraneous, because the base of a logarithm cannot be negative. The other solution, \(\sqrt{14}\), on the other hand, is perfectly valid.
This section demonstrates the solving of logarithmic equations where the bases of the logarithms are different. The formula
$$\log_{a^n}(b^n) = \log_{a}(b)$$
is often very useful for these problems.
Example 3: Solve the equation \(\log_{4}(6x) = \log_{2}(3)\).
Solution: Convert the base of the logarithm on the right-hand side of the equation to \(4\):
$$\log_{4}(6x) = \log_{4}(9)$$
Now remove the logarithms:
$$6x = 9 \Rightarrow$$ $$x = \frac{3}{2}$$
Example 4: Solve this equation for \(y\):
$$\log_{y}(3) = \log_{y^3}(21) - \log_{y}(2)$$
Solution: Add \(\log_{y}(2)\) to both sides of the equation :
$$\log_{y}(3) + \log_{y}(2) = \log_{y^3}(21)$$
Combine the two logarithms on the left-hand side of the equation:
$$\log_{y}(6) = \log_{y^3}(21)$$
Convert the logarithm on the left-hand side to have a base of \(y^3\):
$$\log_{y^3}(216) = \log_{y^3}(21)$$
Take the \(y^3\)rd exponent of both sides of the equation:
$$216 \neq 21$$
Thus there are no solutions.
Example 5: Solve this equation for \(x\): \(\log_{4}(16^{12}) + \log_{16}(x \cdot 4^5) = 13\).
Solution: There are a couple different ways to handle this. Start by converting the exponent inside the first logarithm to one with a base of \(4\):
$$16^{12} = (4 \cdot 4)^12 = 4^{12} \cdot 4^{12} = 4^{12 + 12} = 4^{24}$$
So
$$\log_{4}(4^{24}) + \log_{16}(x \cdot 4^5) = 13 \Rightarrow$$ $$24 + \log_{16}(x \cdot 4^5) = 13 \Rightarrow$$ $$\log_{16}(x \cdot 4^5) = -11$$
Take the \(16\)th exponent of both sides:
$$x \cdot 1024 = 16^{-11} \Rightarrow$$ $$x = \frac{1}{1024 \cdot 16^{11}}$$
Example 6: Solve the equation for \(x\): \(\log_{7}(12) \cdot \log_{3}(49) = 4^{x^3 - 3}\).
Solution: Convert the first logarithm to base \(49\) so the base matches the second logarithm's argument:
$$\log_{49}(144) \cdot \log_{3}(49) = 4^{x^3 - 3}$$
Combine the two logarithms with a logarithm identity:
$$\log_{3}(144) = 4^{x^3 - 3}$$
We can remove factors of \(3\) from this logarithm:
$$\log_{3}(16) + \log_{3}(9) = 4^{x^3 - 3} \Rightarrow$$ $$\log_{3}(16) + \log_{3}(3^2) = 4^{x^3 - 3} \Rightarrow$$ $$\log_{3}(16) + 2 = 4^{x^3 - 3}$$
Now it is just a matter of isolating the variable. Take the base \(4\) logarithm of both sides:
$$\log_{4}(\log_{3}(16) + 2) = x^3 - 3$$
Solve for \(x\):
$$\log_{4}(\log_{3}(16) + 2) + 3 = x^3 \Rightarrow$$ $$x = \sqrt[3]{\log_{4}(\log_{3}(16) + 2)}$$
This set of problems requires exponent laws to be used in addition to the logarithm laws, adding another layer of complexity to these equations.
Example 7: Solve the equation \(49^{36^{x\log_{6}(4) + \log_{36}(2)}} = 49^{14}\).
Solution: Begin by breaking down the double exponent. The two logarithms present can be combined with a combination of three identities:
$$x\log_{6}(4) + \log_{36}(2) =$$ $$x\log_{36}(16) + \log_{36}(2) =$$ $$\log_{36}(16^x) + \log_{36}(2) =$$ $$\log_{36}(16^x \cdot 2)$$
The base of the logarithm is the same as the higher exponent on the left-hand side, so cancel them out:
$$49^{16^x \cdot 2} = 49^{14}$$
Convert the base of the left-hand side to a \(7\), because the base of the right-hand side has two factors of \(7\):
$$(7^{16^x \cdot 2})^2 = (7 \cdot 7)^{14} \Rightarrow$$ $$7^{16^x \cdot 2} \cdot 7^{16^x \cdot 2} = 7^{14} \cdot 7^{14} \Rightarrow$$ $$7^{16^x \cdot 2 + 16^x \cdot 2} = 7^{14 + 14} \Rightarrow$$ $$7^{16^x \cdot 4} = 7^{28}$$
Both lower exponents have the same base. Take the base \(7\) logarithm of both sides:
$$16^x \cdot 4 = 28$$
Isolate the factor of the left-hand side with the variable:
$$16^x = 7$$
Take the base \(16\) logarithm of both sides of the equation:
$$x = \log_{16}(7)$$
This is a rather complex topic, as illustrated by the number of steps needed for Example 7, so we'll give another example.
Example 8: Solve this equation for \(x\): \(9^{x + 3} = 3^{2\log_{2}(56)}\).
Solution: A base conversion would be a good start. Convert the exponent on the left-hand side to one of base \(3\):
$$9^{x + 3} = (3 \cdot 3)^{x + 3} = 3^{x + 3} \cdot 3^{x + 3} = 3^{x + x + 3 + 3} = 3^{2x + 6}$$
The equation then becomes:
$$3^{2x + 6} = 3^{2\log_{2}(56)}$$
Even though it is possible to eliminate the base \(2\) logarithm on the right-hand side, the easiest thing to do is eliminate the exponents entirely with a base \(3\) logarithm:
$$2x + 6 = 2\log_{2}(56)$$
Divide both sides of the equation by \(2\):
$$x + 3 = \log_{2}(56)$$
Isolate \(x\) and simplify the result:
$$x = \log_{2}(56) - 3 = \log_{2}(56) - \log_{2}(8) = \log_{2}(\frac{56}{8}) = \log_{2}(7)$$
This article has demonstrated how logarithm identities and the use of exponents can solve one-variable logarithmic equations. In summary, try combining or breaking apart logarithms based on the bases of the logarithms, the arguments, and prime factorizations of the arguments. Typically a combination of these techniques will give you an answer.
A reference, courtesy of
"The Art of Problem Solving Intermediate Algebra" Rusczyk, Richard and Crawford, Matthew