When Matching Bases Is Not Convenient
Some exponential equations can be solved by rewriting both sides with the same base. That method is especially direct when the numbers have a simple common base. But not every equation has a useful matching-base form. For example, \(5^{2x-1}=13\) does not become simpler by rewriting 5 and 13 as powers of one common base.
Logarithms provide a general method for these cases. Since a logarithm is the inverse of an exponential function, taking a logarithm allows us to bring an exponent down as a factor and then solve for the variable.
The Basic Method
Suppose the equation has the form \(a^{mx+b}=c\), where \(a>0\), \(a\ne1\), \(m\ne0\), and \(c>0\). The base must be positive for the real-valued exponential expression to be defined, and it cannot equal 1 because \(1^{mx+b}\) is always 1. The right side must be positive because \(a^{mx+b}>0\) for every real \(x\).
Take the natural logarithm of both sides. The logarithm power rule gives \(\ln(a^u)=u\ln a\) for \(a>0\). Therefore:
Because \(a\ne1\), \(\ln a\ne0\). We can divide by \(\ln a\), then isolate \(x\):
The same solution results if we use any logarithm base greater than zero and not equal to 1, provided we use that base consistently on both sides. Natural logarithms are a convenient choice. The quotient \(\frac{\ln c}{\ln a}\) expresses the logarithm of \(c\) to base \(a\), but we do not need to change the equation to that notation to solve it.
Why Taking Logarithms Works
The method depends on two properties. First, the logarithm is the inverse of the exponential function: for \(a>0\), \(a\ne1\), and \(y>0\), \(\log_a(a^y)=y\). Second, the logarithm power rule states that \(\ln(u^r)=r\ln u\) when \(u>0\). Here is the full chain of reasoning for the general equation.
Both sides of \(a^{mx+b}=c\) are positive, so taking their natural logarithms is defined and gives an equivalent equation.
\(\ln(a^{mx+b})=(mx+b)\ln a\), so the variable expression in the exponent becomes a factor.
Since \(a\ne1\), \(\ln a\ne0\); since \(m\ne0\), the resulting linear equation can be solved uniquely for \(x\).
This also shows why the conditions matter. If \(c\le0\), the original equation has no real solution because a positive base raised to a real power is always positive. If \(m=0\), the exponent \(mx+b\) does not depend on \(x\): the equation is either true for every real \(x\) or false for every real \(x\), rather than determining one value. The special base \(a=1\) has the same constant-expression issue.
Worked Examples
Worked Example: Solve \(5^{2x-1}=13\)
The bases are not conveniently matched, so take natural logarithms of both sides. The right side is positive, as required.
This exact expression is the solution. Using a calculator and rounding \(x\) to nine decimal places gives \(x\approx1.296846321\). Substitution into the original equation gives an exponent of approximately \(2(1.296846321)-1=1.593692642\), so \(5^{1.593692642}\approx13.0000000\). For an exact check, substitute the unrounded expression:
Thus the exact solution satisfies the original equation.
Worked Example: Solve \(2^{3x+1}=11\)
Take natural logarithms and use the power rule. Then isolate \(x\).
The exact solution is \(x=\frac{\frac{\ln 11}{\ln 2}-1}{3}\). A calculator gives \(x\approx0.819810540\) to nine decimal places. Substituting this rounded value gives \(3x+1\approx3.459431620\), and hence \(2^{3x+1}\approx11.00000001\). The small difference from 11 is due to rounding. With the exact expression, the check is:
Worked Example: Solve \(4(1/3)^{x+2}=5\)
First isolate the exponential expression by dividing both sides by 4. The resulting right side, \(5/4\), is positive, so its logarithm is defined.
Since \(\ln(1/3)<0\), the quotient is negative. Numerically, \(x\approx-2.203114013\), rounded to nine decimal places. Check the exact expression in the original equation:
The negative logarithm of the base causes no difficulty: it is simply a nonzero number by which we divide.
Equations with a Coefficient
An equation may have a constant multiplying the exponential expression, as in the third example. In general, consider \(A a^{mx+b}=C\). If \(A=0\), the left side is zero, so the equation is true for every \(x\) when \(C=0\) and has no solution when \(C\ne0\). If \(A\ne0\), divide by \(A\) first. The quotient \(C/A\) must be positive; if it is zero or negative, there is no real solution. When \(C/A>0\), the equation becomes \(a^{mx+b}=C/A\), and the logarithm method applies if \(a>0\), \(a\ne1\), and \(m\ne0\).
This order is useful: isolate the exponential expression before taking logarithms. Taking the logarithm of a product is possible when its factors are positive, but dividing first keeps the steps clear and makes the required positivity check explicit.
A Common Mistake: Losing the Logarithm of the Base
A frequent error is to take the logarithm of an exponential power but forget the factor \(\ln a\). For example, a mistaken solution to \(2^{x+1}=9\) might be:
The middle step is incorrect. The power rule gives \(\ln(2^{x+1})=(x+1)\ln2\), not just \(x+1\). The corrected work is:
Check the exact answer: \(2^{x+1}=2^{\ln9/\ln2}=e^{\ln9}=9\). By contrast, the incorrect rounded value \(x\approx1.197225\) gives \(2^{x+1}\approx4.59\), not 9. Always keep the logarithm of the base when using the power rule.
Checking and Interpreting a Solution
A solution obtained through logarithms should be checked in the original equation, not only in the transformed equation. The exact checks above use the inverse relationship between logarithms and exponentials. For a decimal answer, substitution should give a value close to the required right side; any small discrepancy should be consistent with the number of digits retained.
For equations of the form \(a^{mx+b}=c\) under the stated conditions, the method produces exactly one real solution. The logarithm is one-to-one, and the resulting equation is linear with nonzero coefficient \(m\ln a\). This conclusion is specific to that form; an equation with additional terms or more complicated expressions may require further analysis.
Logarithms matter here because they turn an exponent containing the unknown into an ordinary linear expression. Matching bases is often quicker when it applies, but logarithms work for a much broader range of positive right sides and positive bases other than 1.
Check Your Understanding
For each question, show the isolation and logarithm steps, and check any proposed solution in the original equation.
- Solve \(3^{x-2}=8\) exactly, then give a decimal approximation rounded to six decimal places.
- Solve \(7^{2x}=4\) using natural logarithms. Why is the quotient involving \(\ln 7\) positive?
- For \(5(2)^{x+1}=3\), explain why dividing by 5 still leaves an equation to which logarithms can be applied.
- Does \(4^{x}=0\) have a real solution? Explain using the possible values of a positive-base exponential.
- In solving \(6^{2x-1}=10\), what factor must appear when the power is brought down by taking natural logarithms?
