Just like one-variable logarithmic equations, systems of equations involving logarithms require the same combination of techniques: logarithmic identities and exponents, which help to rewrite the logarithms in ways to make it easy to solve for the variables. Algebraic manipulations such as substitution and elimination can help to achieve a one-variable equation that can be solved with ease. Then any remaining variables can be solved for more easily.
There are a few crucial identities that are used to transform logarithmic expressions of a base \(a\):
Sum-product: \(\log_{a}(bc) = \log_{a}(b) + \log_{a}(c)\)
Difference-quotient: \(\log_{a}(\frac{b}{c}) = \log_{a}(b) - \log_{a}(c)\)
Exponents: \(\log_{a}(b^c) = c\log_{a}(b)\)
Conversion of Bases: \(\log_{a^n}(b^n) = \log_{a}(b)\)
Often reducing solving one equation for a variable in terms of the other can be useful, as it allows you to work with a one-variable equation. The general concept is the same as linear systems of equations, but there is more algebra required. Logarithmic identities and exponents will be helpful.
Example 1: Solve this system of equations:
$$\log_{5}(x) = \log_{25}(y^2)$$ $$\log_{4}(y^3) = 7 + \log_{16}(x^2)$$
The logarithms' bases are powers of each other, as \(5^2 = 25\) and \(4^2 = 16\). Convert the logarithms with smaller bases to match the base of the other logarithm in the equation with a logarithm identity:
$$\log_{25}(x^2) = \log_{25}(y^2)$$ $$\log_{16}(y^6) = 7 + \log_{16}(x^2)$$
Apply a base \(25\) exponent to the first equation:
$$x^2 = y^2 \Rightarrow$$ $$x = y$$
We cannot include \(x = -y\) because otherwise some of the arguments of the logarithms will inevitably be \(0\) or negative. In other words, we know that \(x\) and \(y\) are both positive, so \(x \neq -y\). Similar arguments are applicable in many such problems. Substitute \(x = y\) into the second equation:
$$\log_{16}(x^6) = 7 + \log_{16}(x^2)$$
Move the logarithms to the same side:
$$\log_{16}(x^6) - \log_{16}(x^2) = 7$$
The logarithms can be combined with an identity since the bases are the same:
$$\log_{16}(x^4) = 7$$
Remove the logarithm:
$$x^4 = 16^7$$
Take the \(4\)th root of both sides:
$$x = 2^7$$
Since \(x = y\), \(y = 2^7\) as well.
Example 2: Solve this system of equations:
$$\log_{6}(8x) = \log_{36}(y^2)$$ $$x^{\log_{6}(4)} = 2^{\log_{36}(y^8)}$$
Solution: Generally, having two expressions in the form of a quotient, where each expression contains one of the variables, makes it hard to create a substitution, so we need to try to avoid creating such a quotient in the first equation. By a logarithm identity:
$$\log_{6}(8x) = \log_{6}(4) \cdot \log_{4}(8x)$$
Divide both sides of the equation by this constant:
$$\log_{4}(8x) = \frac{\log_{36}(y^2)}{\log_{6}(4)}$$
Remove the logarithm on the left-hand side to work on isolating \(x\):
$$8x = 4^{ \frac{\log_{36}(y^2)}{\log_{6}(4)}}$$
Next, divide both sides of the equation by \(8\):
$$x = \frac{4^{ \frac{\log_{36}(y^2)}{\log_{6}(4)}}}{8}$$
Substitute this into the second equation:
$$(\frac{4^{ \frac{\log_{36}(y^2)}{\log_{6}(4)}}}{8})^{\log_{6}(4)} = 2^{\log_{36}(y^8)}$$
This is quite an intimidating equation, but this is one of those times where simplifying the equation one piece at a time can pay off. Rewrite the right-hand side with some logarithm identities:
$$2^{\log_{36}(y^8)} = 2^{8\log_{36}(y)} = 2^{4\log_{36}(y)} \cdot 2^{4\log_{36}(y)} = (2^2)^{4\log_{36}(y)} = 4^{4\log_{36}(y)}$$
So
$$(\frac{4^{ \frac{\log_{36}(y^2)}{\log_{6}(4)}}}{8})^{\log_{6}(4)} = 4^{4\log_{36}(y)}$$
Now the exponents have the same base. Multiply both sides of the equation by \(8^{\log_{6}(4)}\):
$$(4^{\frac{\log_{36}(y^2)}{\log_{6}(4)}})^{\log_{6}(4)} = 4^{4\log_{36}(y)} \cdot 8^{\log_{6}(4)}$$
Take the base 4 logarithm of both sides:
The expression \(8^{\log_{6}(4)}\) should be converted to a base \(4\) logarithm using logarithm identities:
$$8^{\log_{6}(4)} = (2 \cdot 2 \cdot 2)^{\log_{6}(4)} = (2^3)^{\log_{6}(4)} = ((2^2)^{\frac{3}{2}})^{\log_{6}(4)} = (4^{\frac{3}{2}})^{\log_{6}(4)} = 4^{\frac{3}{2}\log_{6}(4)} =$$ $$4^{\log_{6}(4^{\frac{3}{2}})} = 4^{\log_{6}(8)}$$
Therefore:
$$(4^{\frac{\log_{36}(y^2)}{\log_{6}(4)}})^{\log_{6}(4)} = 4^{4\log_{36}(y)} \cdot 4^{\log_{6}(8)} \Rightarrow$$ $$(4^{\frac{\log_{36}(y^2)}{\log_{6}(4)}})^{\log_{6}(4)} = 4^{4\log_{36}(y) + \log_{6}(8)}$$
We can further condense the right-hand side with two more logarithmic identities:
$$(4^{\frac{\log_{36}(y^2)}{\log_{6}(4)}})^{\log_{6}(4)} = 4^{4\log_{36}(y) + \log_{36}(64)} \Rightarrow$$ $$(4^{\frac{\log_{36}(y^2)}{\log_{6}(4)}})^{\log_{6}(4)} = 4^{4\log_{36}(64y)}$$
Take the base \(4\) logarithm of both sides:
$$(\frac{\log_{36}(y^2)}{\log_{6}(4)}) \cdot {\log_{6}(4)} = 4\log_{36}(64y)$$
Some things cancel on the left-hand side:
$$\log_{36}(y^2) - \log_{4}(8) = 4\log_{36}(64y)$$
Move the logarithms with the variable to one side and everything else to the other:
$$\log_{36}(y^2) - 4\log_{36}(64y) = \log_{4}(8)$$
Condense the left-hand side into one logarithm with the use of identities:
$$\log_{36}(y^2) - \log_{36}((64y)^4) = \log_{4}(8) \Rightarrow$$ $$\log_{36}(\frac{1}{64^4y^2}) = \log_{4}(8)$$
Isolate the variable with a base \(36\) exponent and then some algebraic manipulation:
$$\frac{1}{64^4y^2} = 36^{\log_{4}(8)} \Rightarrow$$ $$64^4y^2 = \frac{1}{36^{\log_{4}(8)}} \Rightarrow$$ $$y^2 = \frac{1}{64^4 \cdot 36^{\log_{4}(8)}} \Rightarrow$$ $$y = \frac{1}{8^4 \cdot 6^{\log_{4}(8)}}$$
Substitute this into the first original equation:
$$\log_{6}(8x) = \log_{36}((\frac{1}{8^4 \cdot 6^{\log_{4}(8)}})^2)$$
Convert the base \(36\) logarithm to base \(6\) with an identity:
$$\log_{6}(8x) = \log_{6}(\frac{1}{8^4 \cdot 6^{\log_{4}(8)}})$$
Take the base \(6\) exponent of both sides and then divide both sides of the equation by \(8\):
$$x = \frac{1}{8^5 \cdot 6^{\log_{4}(8)}}$$
Just like in linear systems of equations, it is often possible to set up the equations so that adding/subtracting/multiplying/dividing the two equations will get rid of one variable and leave you with a one-variable equation. Two examples of this will be shown.
Example 3: Solve the system of equations:
$$\log_{7}(x) + \log_{49}(y^4) = 12$$ $$\log_{7}(xy^3) = 6$$
Solution: Convert the base \(49\) logarithm to base \(7\) with an identity:
$$\log_{7}(x) + \log_{7}(y^2) = 12$$
Break up the product in the second equation by using a logarithm identity:
$$\log_{7}(x) + \log_{7}(y^3) = 6$$
This configuration was chosen in order to set up an elimination. Subtract the first equation from the second:
$$\log_{7}(y^3) - \log_{7}(y^2) = -6$$
The logarithms have the same base, so divide the arguments with another logarithm identity:
$$\log_{7}(y) = -6$$
Use an exponent to isolate \(y\):
$$y = 7^{-6}$$
Substitute into the second original equation:
$$\log_{7}(x \cdot 7^{-6}) = 6$$
Break up this product and create a sum, just like we did earlier:
$$\log_{7}(x) + \log_{7}(7^{-6}) = 6$$
The second logarithm can be simplified, since exponents and logarithms are inverse operations:
$$\log_{7}(x) - 6 = 6 \Rightarrow$$ $$\log_{7}(x) = 12$$
Now use a base \(7\) exponent to solve for \(x\):
$$x = 7^{12}$$
Example 4: Solve the system of equations:
$$\log_{5}(x^2) + \log_{125}(y^3) = 5$$ $$\log_{25}(x^2y^2) = 3$$
Solution: We see that \(125 = 5^3\), so we can transform the second logarithm in the first equation to one of base \(5\):
$$\log_{5}(x^2) + \log_{5}(y) = 5$$
Now these two logarithms have the same base. Combine them by multiplying the arguments, valid by a logarithm identity:
$$\log_{5}(x^2y) = 5$$
This logarithm is in base \(5\), so convert the logarithm in the second equation to base \(5\) to make setting up an elimination of one of the variables easier:
$$\log_{5}(xy) = 3$$
Both equations now have only base \(5\) logarithms. We can convert this to an exponential system by taking the base \(5\) exponent of both sides of both equations.
$$x^2y = 5^5$$ $$xy = 5^3$$
Now divide the second equation from the first to eliminate \(y\):
$$x = 5^{5 - 3} = 5^2 = 25$$
Substitute this into the second original equation:
$$\log_{25}(y^2 \cdot 25^2) = 3$$
This logarithm can be broken up into a sum with the sum-product logarithm identity:
$$\log_{25}(y^2) + \log_{25}(25^2) = 3$$
The second logarithm can be removed since the exponent has the same base:
$$\log_{25}(y^2) + 2 = 3 \Rightarrow$$ $$\log_{25}(y^2) = 1$$
Remove the logarithm with a base \(25\) exponent and solve for \(y\):
$$y^2 = 25 \Rightarrow$$ $$y = \pm 5$$
Finally, notice that using \(x = -5\) does not make any individual expressions in the original equations undefined. Often using negative values of variables as solutions to logarithmic equations creates a problem here, but in this case it does not, because every instance of \(x\) in the original equations includes an even exponent, removing the negative signs.
This section details how certain logarithmic systems of equations can be written as two-variable linear systems, which makes them much easier to solve. Typically a substitution is required here.
Example 5: Solve the system of equations:
$$4\log_{2}(8x^3) + \log_{5}(y^6) = 17$$ $$\log_{2}(x^5) + \log_{5}(y^2) = 3$$
Solution: To transform to a linear system, we need to make the logarithmic expressions look alike with identities. Later, we will set \(a\) and \(b\) equal to logarithmic expressions involving \(x\) and \(y\), respectively. Dealing with \(y\) first, notice that
$$y^6 = y^{2 + 2 + 2} = y^{2} \cdot y^2 \cdot y^2 = (y^2)^3$$
So using logarithm identities:
$$4\log_{2}(8x^3) + \log_{5}((y^2)^3) = 17 \Rightarrow$$ $$4\log_{2}(8x^3) + 3\log_{5}(y^2) = 17$$
The logarithms involving \(x\) require more work. In the second equation, remove the exponent from the logarithm with the \(x\) using an identity:
$$5\log_{2}(x) + \log_{5}(y^2) = 3$$
As for the first equation, we must use a combination of exponential and logarithmic identities to create an expression involving \(\log_{2}(x)\):
Step 1: Remove the exponent. $$4\log_{2}(8x^3) = 4\log_{2}(2 \cdot 2 \cdot 2 \cdot x \cdot x \cdot x) = 4\log_{2}(2x \cdot 2x \cdot 2x) = 4\log_{2}((2x)^3) =$$ $$4 \cdot 3\log_{2}(2x) = 12\log_{2}(2x)$$
Step 2: Remove the constant from the argument of the logarithm:
$$12\log_{2}(2x) = 12(\log_{2}(2) + \log_{2}(x)) = 12(1 + \log_{2}(x)) = 12 + 12\log_{2}(x)$$
Therefore the system becomes
$$12 + 12\log_{2}(x) + 3\log_{5}(y^2) = 17$$ $$5\log_{2}(x) + \log_{5}(y^2) = 3$$ $$\Downarrow$$ $$12\log_{2}(x) + 3\log_{5}(y^2) = 5$$ $$5\log_{2}(x) + \log_{5}(y^2) = 3$$
We can now substitute: \(a = \log_{2}(x)\) and \(b = \log_{5}(y^2)\).
$$12a + 3b = 5$$ $$5a + b = 3$$
This is a linear system which can be solved easily. Multiply the second equation by \(3\):
$$15a + 3b = 9$$
Subtract the first equation from this to eliminate \(b\):
$$3a = 4 \Rightarrow$$ $$a = \frac{4}{3}$$
Therefore,
$$\log_{2}(x) = \frac{4}{3}$$
Use a base \(2\) exponent to immediately solve for \(x\):
$$x = 2^{\frac{4}{3}}$$
Now use the value of \(a\) to find \(b\). Substitute into the second original equation:
$$5 \cdot \frac{4}{3} + b = 3$$
Solve for \(b\):
$$\frac{20}{3} + b = \frac{9}{3} \Rightarrow$$ $$b = -\frac{11}{3}$$
Solve for \(y\):
$$-\frac{11}{3} = \log_{5}(y^2)$$
Use a base \(5\) exponent:
$$5^{-\frac{11}{3}} = y^2$$
Take the square root:
$$y = \pm 5^{-\frac{11}{6}}$$
Example 6: Show that this system has no solutions:
$$\log_{6}(x^2) + 4\log_{9}(y^2) = 4$$ $$\log_{36}(x^2) + \log_{81}(y^8) = 7$$
Solution: \(6^2 = 36\) and \(9^2 = 81\), so changing the bases of the two logarithms in the first equation respectively can aid in a transformation. Use a logarithm identity and square the arguments:
$$\log_{36}(x^4) + 4\log_{81}(y^4) = 4$$ $$\log_{36}(x^2) + \log_{81}(y^8) = 7$$
We now need to match up the terms with the logarithms. Notice we can change some of the arguments with a logarithm identity:
$$\log_{36}(x^4) = 2\log_{36}(x^2)$$ $$\log_{81}(y^8) = 2\log_{81}(y^4)$$
Therefore,
$$2\log_{36}(x^2) + 4\log_{81}(y^4) = 7$$ $$\log_{36}(x^2) + 2\log_{81}(y^4) = 4$$
Now we can transform this into a linear system by substituting \(a = \log_{36}(x^2)\) and \(b = \log_{81}(y^4)\):
$$2a + 4b = 7$$ $$a + 2b = 4$$
To prove this system has no solutions, divide the first equation by \(2\):
$$a + 2b = \frac{7}{2}$$
Thus the system claims that \(\frac{7}{2} = 4\), which is not true, so the system has no solutions.
Similarly to exponential systems of equations, logarithmic systems of equations can be manipulated using the central principles of exponents and logarithms, particularly identities, to create equations that are easy to solve, either a simple one-variable logarithmic or exponential equation, or a system of linear equations. Often testing different manipulations on a system (e.g. applying three different identities to an equation) can create an indication as to which technique to use. More complicated systems usually require multiple identities and possibly more than one substitution.
Reference, courtesy of
"The Art of Problem Solving Intermediate Algebra" Rusczyk, Richard and Crawford, Matthew