Like expressions, equations can contain exponents all the same. Using the laws of exponents, it is possible to rewrite either side of the equation to make it easier to solve. In some cases, it is helpful to use logarithms. The logarithm is defined as the inverse operation of an exponent, as the two operations undo each other. Every logarithm has a base, which indicates what effect the logarithm has on an exponential expression placed inside it. In general:
If \(a > 0\):
$$\log_{a}(a^b) = b$$ $$a^{\log_a(b)} = b$$
However this article will focus on the first identity shown above. Thus it can be seen that logarithms and exponents are inverse operations--they undo each other, just like addition and subtraction or multiplication and division.
There are some instances where logarithms are not necessary, or it is intuitively obvious what should be done. These problems often use division or multiplication. Multiple examples will be given.
Example 1: Solve this equation for \(x\): \(2^{4x} = 4^{3x + 7}\).
Solution: Notice that \(4 = 2^2\), so the right-hand side can be written with a base of \(2\):
$$4^{3x + 7} = 2^{3x + 7} \cdot 2^{3x + 7} = 2^{3x + 3x + 7 + 7} = 2^{2(3x + 7)} = 2^{6x + 14}$$
This gives the equation
$$2^{4x} = 2^{6x + 14}$$
Now since the bases are the same, the superscripts must also be the same for the two sides of the equation to be equivalent. Consider a graph of \(2^x\), and it is always increasing as \(x\) increases (this is called a monotonically increasing function). All exponential functions with a base greater than \(1\) have this pattern.
Therefore:
$$4x = 6x + 14$$
Solve this by isolating \(x\):
$$-2x = 14 \Rightarrow$$ $$x = -7$$
The step where we transformed the exponential equation to a linear equation could also be done with a base 2 logarithm.
Example 2: Prove that \(4^{7x} = 64^{3x}\) has no positive solutions.
Solution: Although this is a slightly different type of problem than the previous one, similar concepts are still applicable. \(64 = 4 \cdot 4 \cdot 4\), so the base of the right-hand side can be factored:
$$4^{7x} = 4^{3x} \cdot 4^{3x} \cdot 4^{3x} \Rightarrow$$ $$4^{7x} = 4^{9x}$$
The superscripts must be equivalent:
$$7x = 9x$$
The only solution to this equation is \(x = 0\), so there are no positive solutions.
This last example of solving an exponential equation without logarithms requires simplification.
Example 3: Solve for \(x\): $$\frac{117^{x + 2}}{3^{2x + 3}} = 13^{x + 4}$$.
Solution: \(117\) is rather large for a base, but observe its factorization:
$$117 = 3 \cdot 3 \cdot 13$$
This is extremely helpful, because a lot of simplifying can be done. Factor the base in the equation:
$$\frac{3^{x + 2} \cdot 3^{x + 2} \cdot 13^{x + 2}}{3^{2x + 3}} = 13^{x + 4}$$
First off, divide both sides of the equation by \(13^{x + 2}\):
$$\frac{3^{x + 2} \cdot 3^{x + 2}}{3^{2x + 3}} = 13^2 = 169$$
All the exponents on the left side have the same base, so they can all be combined at once, adding the superscripts in the numerator and then subtracting the superscript of the denominator from that:
$$3^{x + 2 + x + 2 - 2x - 3} = 169 \Rightarrow$$ $$3^{2x - 2x + 4 - 3} = 169 \Rightarrow$$ $$3^{1} = 169 \Rightarrow$$ $$3 \neq 169$$
As a result, this equation has no solutions.
The phrase "common logarithm" refers to the logarithm with a base of 10, because it is very commonly used in real-world applications. For example, the common logarithm can be used to find the pH, or concentration of hydrogen with a +1 charge, of a chemical. The common logarithm is denoted \(\log_{10}\).
Example 4: The common logarithm is helpful when an expression is in scientific notation, because the exponent's base will be \(10\). Many numbers in word problems in physics and chemistry will be written in scientific notation.
Logarithms can be used to solve exponential equations by undoing the effect of an exponential equation. However, when the bases of exponents are not the same, using logarithms becomes less straightforward. However, exponent laws are often helpful to make the usage if logarithms more manageable.
Example 5: Solve this equation for \(x\): \(44^{2x} = 256 \cdot 11^{2x + 1}\).
Solution: A common necessary technique is to factor a base. 11 is a factor of 44, so factoring the 44 base will be helpful for simplifying the equation:
$$4^{2x} \cdot 11^{2x} = 256 \cdot 11^{2x + 1}$$
Now divide both sides of the equation by \(11^{2x}\):
$$4^{2x} = 256 \cdot 11 = 2816$$
This is where the logarithms come into play. Take the base 4 logarithm of both sides of the equation:
$$2x = \log_{4}(2816)$$
Finally divide both sides of the equation by \(2\):
$$x = \frac{\log_{4}(2816)}{2}$$
This is a more complicated example with more exponents.
Example 6: Solve this equation for \(x\): \(6^x \cdot 7^{2x} = 3^{x + 1} \cdot 14^{x - 4}\).
Solution: The trick here is to notice that the bases all factor into 2s, 3s, and 7s. Expand both sides so that all bases are one of these three numbers (factor the 6 and 14):
$$2^x \cdot 3^x \cdot 7^{2x} = 3^{x + 1} \cdot 2^{x - 4} \cdot 7^{x - 4}$$
While this appears messy, we can divide both sides of the equation by exponential expressions to make it easier to manipulate. Begin by dividing both sides of the equation by \(2^{x - 4}\):
$$2^4 \cdot 3^x \cdot 7^{2x} = 3^{x + 1} \cdot 7^{x - 4}$$
Now divide both sides of the equation by \(3^x\):
$$2^4 \cdot 7^{2x} = 3 \cdot 7^{x - 4}$$
Finally divide both sides of the equation by \(7^{x - 4}\):
$$16 \cdot 7^{x + 4} = 3$$
Move all constants to the right side:
$$7^{x + 4} = \frac{3}{16}$$
The fact that operations were done to maintain positive exponents made the algebra easier. Now take the base 7 logarithm of both sides of the equation:
$$x + 4 = \log_{7}\left(\frac{3}{16}\right)$$
Finally subtract \(4\) from both sides of the equation to get a value for \(x\):
$$x = \log_7\left(\frac{3}{16}\right) - 4$$
Here is an example where one side of the equation has a fraction.
Example 7: Solve this equation for \(z\): \(\frac{16^z}{8^z} = 2^{2z - 7}\).
Solution: Even with a fraction, this problem is not very difficult because the numerator and denominator have the same superscript. That means we can rewrite the left side as follows:
$$\frac{16^z}{8^z} = \left(\frac{16}{8}\right)^z = 2^z$$
Then the equation becomes
$$2^z = 2^{2z - 7}$$
Here we can use the base 2 logarithm on both sides:
$$z = 2z - 7$$
Finally, solve this linear equation for \(z\):
$$-z = -7 \Rightarrow$$ $$z = 7$$
Here is one final example of using logarithms. This one requires the use of logarithms twice.
Example 8: Solve this equation for \(x\): \(2^{15^{4x}} = 2^{5^{4x}}\).
Solution: The double exponent makes this equation look intimidating, but the basic rules of logarithms will still apply, regardless how complicated the problem itself is. We can look at the left side having a base of \(2\) and a superscript of \(15^{4x}\), while looking at the right side as having a base of \(2\) and a superscript of \(5^{4x}\). Thus we can take the base 2 logarithm of both sides:
$$15^{4x} = 5^{4x}$$
This is a much simpler problem, as there are no longer any double exponents. Since \(5\) is a factor of \(15\), factor the base on the left-hand side:
$$5^{4x} \cdot 3^{4x} = 5^{4x}$$
Divide both sides of the equation by \(5^{4x}\):
$$3^{4x} = 1$$
Now we use logarithms again, but this time we use the base 3 logarithm:
$$4x = \log_{3}(1)$$
Notice that \(3^0 = 1\), and as a result \(\log_3{1} = 0\), by taking the base 3 logarithm of both sides. Thus:
$$4x = 0 \Rightarrow$$ $$x = 0$$
By introducing properties of logarithms, the number of exponential equations that you should now be able to solve have greatly increased. The exponent laws are also very helpful, because they allow you to transform expressions into forms where logarithms can be used to reduce an exponential equation to a linear equation. Obviously a wide variety of problems exist, but the same tools are useful for this great spectrum, of which only a small amount was shown here.