Typically, exponential equations require one or more logarithms to solve. In many cases, the techniques to solve systems of equations and the laws of exponents must be combined to solve exponential systems of equations. Treating the system like a basic linear system of equations is sometimes a helpful tool as well.
By "solving by substitution," we mean that these problems involve making a substitution to reduce an equation to one variable. It is basically the same method as substituting expressions in linear systems, except more algebra is usually required to fully solve the system.
Example 1: Solve this system for \(x\) and \(y\):
$$6^y = 36^{x + 4}$$ $$36^y = 36^{x + 6}$$
Solution 1: There are a few different substitutions we can make with a little manipulation. However, observe that \((6^y)^2 = 36^y\), so square the first equation:
$$(6^y)^2 = (36^{x + 4})^2 \Rightarrow$$ $$6^y \cdot 6^y = 36^{x + 4} \cdot 36^{x + 4} \Rightarrow$$ $$(6^2)^y = 36^{x + 4 + x + 4} \Rightarrow$$ $$36^y = 36^{2x + 8}$$
Now we can solve for \(x\) by substituting \(36^y = 36^{x + 6}\) into this equation:
$$36^{x + 6} = 36^{2x + 8}$$
We can begin by taking the base \(36\) logarithm of both sides of the equation:
$$x + 6 = 2x + 8$$
Now solve this linear equation:
$$6 = x + 8 \Rightarrow$$ $$x = -2$$
To solve for \(y\), substitute this value of \(x\) into either original equation. We will use the second original equation since both exponents have the same base:
$$36^{y} = 36^{x + 6} \Rightarrow$$ $$36^{y} = 36^{-2 + 6} \Rightarrow$$ $$36^{y} = 36^{4}$$
Taking the base \(36\) logarithm of both sides gives \(y = 4\).
Solution 2: This solution illustrates another substitution that can be made. Transform the right-hand side of the second equation to match the right-hand side of the first equation:
$$36^y = 36^{x + 6} \Rightarrow$$ $$36^{y – 2} = 36^{x + 4}$$
Thus we can substitute to get an equation only involving \(y\):
$$36^{y – 2} = 6^y$$
Convert the left-hand side to have a base of \(6\):
$$36^{y – 2} = (6^{y – 2})^2 = 6^{y – 2} \cdot 6^{y – 2} = 6^{y – 2 + y – 2} = 6^{2y – 4}$$
So
$$6^{2y – 4} = 6^y \Rightarrow$$ $$2y – 4 = y \Rightarrow$$ $$y = 4$$
Plug this into the second original equation:
$$36^{4} = 36^{x + 6}$$
Solve for \(x\) with a base \(36\) logarithm:
$$4 = x + 6 \Rightarrow$$ $$x = -2$$
Example 2: Solve this system of equations for\(x\) and \(y\):
$$7^{2x + 4} = 7^{y - 3}$$ $$y - x = 2$$
Solution: It is easier to solve the linear equation for one variable in terms of the other and substitute rather than the other way around.
$$y - x = 2 \Rightarrow$$ $$y - 2 = x$$
Instead of substituting this in for \(x\), we can make a substitution just for \(2x + 4\). Therefore we want to solve for \(2x + 4\) in terms of \(y\) rather than \(x\). Add \(2\) to both sides of the equation:
$$y = x + 2 \Rightarrow$$
Now multiply both sides of the equation by \(2\):
$$2y = 2x + 4$$
Therefore,
$$7^{2y} = 7^{y - 3}$$
Take the base \(7\) logarithm of both sides of the equation:
$$2y = y - 3$$
Solve for \(y\):
$$y = -3$$
Now solve for \(x\) via substitution:
$$-3 - x = 2 \Rightarrow$$ $$3 + x = -2 \Rightarrow$$ $$x = -5$$
Eliminating a variable is a less common technique for exponential systems than substitution, but it still has uses, often resembling its applications in linear systems.
Example 3: Solve this system of equations:
$$6^{x + 2} = 36 \cdot 12^{y + 5}$$ $$6^x = 12^{2y - 11}$$
Solution: \(x\) will be much easier to eliminate because \(6^{x + 2} = 6^x \cdot 36\). Divide both sides of the first equation by \(36\):
$$6^x = 12^{y + 5}$$
Subtract the second equation from the first:
$$12^{y + 5} - 12^{2y - 11} = 0 \Rightarrow$$ $$12^{y + 5} = 12^{2y - 11}$$
Take the base \(12\) logarithm of both sides:
$$y + 5 = 2y - 11$$
Solve for \(y\):
$$5 = y - 11 \Rightarrow$$ $$y = 16$$
Plug this value into the second original equation:
$$6^x = 12^{2(16) - 11} \Rightarrow$$ $$6^x = 12^{21}$$
We must factor the base on the right-hand side:
$$6^x = 2^{21} \cdot 6^{21}$$
Move all the base \(6\) exponents to one side:
$$6^{x - 21} = 2^{21}$$
Take the base \(6\) logarithm of both sides of the equation and isolate \(x\):
$$x - 21 = \log_{6}(2^{21}) \Rightarrow$$ $$x = 21\log_{6}(2) + 21$$
Example 4: Solve this system of equations:
$$4^{2x} + 7^{y – 4} = 305$$ $$2^{4x + 1} + 7^{y - 6} = 513$$
Solution: Convert the left-hand side of the first equation to have exponents of bases \(2\) and \(7\):
$$2^{4x} + 7^{y – 4} = 305$$
Now, notice that the base \(2\) exponents in the two equations have superscripts that differ by \(1\). Multiply the first equation by \(2\) to make these superscripts the same:
$$2^{4x + 1} + 2 \cdot 7^{y – 4} = 610$$
Trying to eliminate \(x\) now will just result in an equation that cannot be solved for \(y\) with algebraic methods. We can match the other exponents by multiplying the second original equation by \(7^2 = 49\). This gives us the system
$$2^{4x + 1} + 2 \cdot 7^{y – 4} = 610$$ $$49 \cdot 2^{4x + 1} + 7^{y – 4} = 25137$$
Although this problem uses elimination, we should use substitutions here before continuing. Let \(a = 2^{4x + 1}\) and \(b = 7^{y – 4}\). Then the system becomes
$$a + 2b = 610$$ $$49a + b = 25137$$
Now we can choose a variable to eliminate. Multiply the second equation by \(2\):
$$98a + 2b = 50274$$
Subtract the first equation from this one to eliminate \(b\):
$$97a = 49664$$
Solve for \(a\):
$$a = 512$$
Substitute this into the first equation from the linear system and solve for \(b\):
$$512 + 2b = 610 \Rightarrow$$ $$2b = 98 \Rightarrow$$ $$b = 49$$
We now undo our substitutions to get \(x\) and \(y\):
$$2^{4x + 1} = 512$$ $$7^{y – 4} = 49$$
Cancel out the exponents with the appropriate logarithms, those of bases \(2\) and \(7\), respectively:
$$4x + 1 = 9$$ $$y – 4 = 2$$
Solve these linear equations for \(x\) and \(y\):
$$4x = 8 \Rightarrow$$ $$x = 2$$
$$y = 6$$
Sometimes, a good solution is to convert both equations to linear equations by using exponent laws and then canceling out all the exponents at once with a logarithm. Generally, the initial exponents' bases will either all be identical, or one is a perfect square, cube, etc. of the other. Here is an example of that procedure.
Example 5: Solve this system for \(x\) and \(y\):
$$4^{2x} \cdot 16^{x + 2} = 16^y$$ $$16^{2x} = 4^{2y + 21}$$
Solution: Convert the left-hand side of the first equation entirely to base \(4\):
$$4^{2x} \cdot (4^2)^{x + 2} = 16^y \Rightarrow$$ $$4^{2x} \cdot 4^{x + 2} \cdot 4^{x + 2} = 16^y \Rightarrow$$ $$4^{2x + x + 2 + x + 2} = 16^y \Rightarrow$$ $$4^{4x + 4} = 16^y$$
Now convert the left-hand side back to base \(16\):
$$16^{2x + 2} = 16^y$$
Take the base \(16\) logarithm of both sides:
$$2x + 2 = y$$
We now look at the next equation:
$$16^{2x} = 4^{2y + 21}$$
Now we can convert the left-hand side to a base \(4\) exponent:
$$(4^2)^{2x} = 4^{2y + 21} \Rightarrow$$ $$4^{2x} \cdot 4^{2x} = 4^{2y + 21} \Rightarrow$$ $$4^{4x} = 4^{2y + 21}$$
Apply the base \(4\) logarithm:
$$4x = 2y + 21$$
So the linear system is
$$4x = 2y + 21$$ $$2x + 2 = y$$
Substitute the value of \(y\) given in terms of \(x\) directly into the first equation:
$$4x = 2(2x + 2) + 21 \Rightarrow$$ $$4x = 4x + 4 + 21 \Rightarrow$$ $$0 \neq 25$$
Therefore there are no solutions.
Example 6: Solve this system of equations:
$$25^{2x + y} = 15625$$ $$25^x \cdot 5^{2y} = 625$$
Solution: In order for all exponents to be of base \(25\), notice that
$$5^{2y} = (5^2)^y = 25^y$$
So combining this with the other exponent in that equation:
$$25^{2x + y} = 15625$$ $$25^{x + y} = 625$$
We can now eliminate a variable, namely \(y\). However, rather than subtracting the equations, we must divide one from the other to eliminate \(y\):
$$\frac{25^{2x + y}}{25^{x + y}} = 25 \Rightarrow$$ $$25^{2x - x + y - y} = 25 \Rightarrow$$ $$25^x = 25$$
Taking the base \(25\) logarithm of both sides of the equation gives \(x = 1\). Plug this into the equation\(25^{x + y} = 625\):
$$25^{1 + y} = 625$$
Take the base \(25\) logarithm of both sides:
$$1 + y = 2 \Rightarrow$$ $$y = 1$$
Exponential systems of equations are more difficult counterparts of linear systems, because they require additional exponent laws and logarithm usage to solve. However it can be seen that, with those added complexities aside, the two types of systems are very similar and can be treated rather similarly. Sometimes an exponential system can be manipulated in such a way that two substitutions can transform the system into a linear system, and sometimes the equations can be manipulated to eliminate a variable and then just solve for one variable. There are many different techniques to try, and every problem will have a different string of manipulations required.
A reference, courtesy of
"The Art of Problem Solving Intermediate Algebra" Rusczyk, Richard and Crawford, Matthew