How do you solve an exponential equation with a base other than e, such as 5^x = 40?

Take ln (or log) of both sides and use the power law ln of 5 to the power x equals x times ln 5. So x times ln 5 equals ln 40, giving x equals ln 40 divided by ln 5, which is the same as log base 5 of 40, about 2.29.

Why do you have to check the domain when solving log equations?

A logarithm is only defined for a positive argument, so after solving you must substitute each answer back and reject any that make the argument of a log zero or negative. For log x plus log of x minus 3 equals 1 the quadratic gives x equals 5 or x equals minus 2, but x equals minus 2 makes log of a negative number, so it is rejected and x equals 5.

How do you solve an equation where the exponent is linear, like e^(2x-1) = 7?

Take ln of the whole side first so the exponent comes down: 2x minus 1 equals ln 7. Then solve the linear equation, giving x equals (1 plus ln 7) divided by 2, about 1.47. Do not just write x equals ln 7.

How do you solve an equation that is a quadratic in disguise, like 2^(2x) - 5(2^x) + 4 = 0?

Substitute y equals 2 to the power x. Since 2 to the power 2x equals (2 to the power x) squared, the equation becomes y squared minus 5y plus 4 equals 0, which factorises to (y minus 1)(y minus 4) equals 0. Then 2 to the power x equals 1 gives x equals 0, and 2 to the power x equals 4 gives x equals 2.

How do you solve an exponential equation with a base other than e, such as 5^x = 40?

Take ln (or log) of both sides and use the power law ln of 5 to the power x equals x times ln 5. So x times ln 5 equals ln 40, giving x equals ln 40 divided by ln 5, which is the same as log base 5 of 40, about 2.29.

Why do you have to check the domain when solving log equations?

A logarithm is only defined for a positive argument, so after solving you must substitute each answer back and reject any that make the argument of a log zero or negative. For log x plus log of x minus 3 equals 1 the quadratic gives x equals 5 or x equals minus 2, but x equals minus 2 makes log of a negative number, so it is rejected and x equals 5.

How do you solve an equation where the exponent is linear, like e^(2x-1) = 7?

Take ln of the whole side first so the exponent comes down: 2x minus 1 equals ln 7. Then solve the linear equation, giving x equals (1 plus ln 7) divided by 2, about 1.47. Do not just write x equals ln 7.

How do you solve an equation that is a quadratic in disguise, like 2^(2x) - 5(2^x) + 4 = 0?

Substitute y equals 2 to the power x. Since 2 to the power 2x equals (2 to the power x) squared, the equation becomes y squared minus 5y plus 4 equals 0, which factorises to (y minus 1)(y minus 4) equals 0. Then 2 to the power x equals 1 gives x equals 0, and 2 to the power x equals 4 gives x equals 2.

Year 12 Methods (Unit 3 & 4) Exponential and logarithmic functions

Solving equations involving exponential and logarithmic functions

Q: How do you solve an exponential equation like e^x = 7?

Take the natural logarithm of both sides, because ln is the inverse of e to the power x. For e to the power x equals 7 this gives x equals ln 7, the exact answer, which is about 1.95 by technology.

Q: How do you solve a logarithmic equation like log_2 x = 5?

Rewrite the logarithm in index form, since log base a of x equals k is the same as x equals a to the power k. So log base 2 of x equals 5 gives x equals 2 to the power 5, which is 32. Similarly ln x equals 3 gives x equals e cubed.

20 practice questions 0 video lessons Theory + worked examples

In Year 12 Mathematical Methods (Queensland, QCAA), you solve exponential equations by taking a logarithm of both sides (\(e^x=k\Rightarrow x=\ln k\)) and logarithmic equations by rewriting them in index form (\(\log_a x=k\Rightarrow x=a^k\)). The log laws combine several logs into one, and every log solution is checked against the domain — a log needs a positive argument, so out-of-range answers are rejected.

An exponential equation has the unknown in the exponent (like \(e^x=7\) or \(5^x=40\)); a logarithmic equation has the unknown inside a logarithm (like \(\ln x=3\) or \(\log_2 x=5\)). Because \(y=e^x\) and \(y=\ln x\) are inverses, and \(a^x=b\) is the same statement as \(x=\log_a b\), each type is undone by the other.

To solve an exponential equation, take a logarithm of both sides; the power law \(\log a^x=x\log a\) then brings the unknown down out of the exponent. To solve a logarithmic equation, rewrite in index form: \(\log_a x=k\) becomes \(x=a^k\). When several logs appear, first combine them with the log laws into a single logarithm.

Logarithms are only defined for a positive argument, so a log equation can produce answers that must be rejected. Always substitute each solution back: if it makes the argument of any log zero or negative, discard it. Some equations are quadratics in disguise — a substitution such as \(y=2^x\) or \(m=\ln x\) turns them into a quadratic you can factorise.

Key idea. Exponential: take \(\ln\), so \(e^x=k\Rightarrow x=\ln k\) and \(r^x=k\Rightarrow x=\dfrac{\ln k}{\ln r}\). Logarithmic: use index form, so \(\log_a x=k\Rightarrow x=a^k\). Then check the domain.

Solve \(e^x=5\): meet \(y=5\) at \(x=\ln 5\approx 1.61\)

Solve \(\ln x=1\): meet \(y=1\) at \(x=e\approx 2.72\)

Solving an exponential equation (take a logarithm; \(r>0\), \(k>0\)):

\[r^{x}=k \;\Rightarrow\; x=\frac{\ln k}{\ln r}=\log_{r}k \qquad e^{x}=k \;\Rightarrow\; x=\ln k\]

e^{x} = k \Rightarrow x = \ln k

Solving a logarithmic equation (rewrite in index form):

\[\log_{a}x=k \;\Leftrightarrow\; x=a^{k} \qquad (a^{x}=b \;\Leftrightarrow\; x=\log_{a}b)\]

\log_{a} x = k \Leftrightarrow x = a^{k}

The logarithm laws used to combine logs before solving:

\[\log_a x+\log_a y=\log_a(xy),\quad \log_a x-\log_a y=\log_a\tfrac{x}{y},\quad \log_a x^{n}=n\log_a x\]

\log_{a} x + \log_{a} y = \log_{a} (x y)

Check the domain. A logarithm needs a positive argument. After solving, discard any \(x\) that makes an argument \(\le 0\) — e.g. \(\log x+\log(x-3)=1\) gives \(x=5\) or \(x=-2\), but only \(x=5\) satisfies \(x>3\).

How to solve a logarithmic equation

Combine the logs. Use the sum, difference and power laws to write one logarithm on each side (or a single \(\log = \text{number}\)).
Undo the log. Rewrite \(\log_a(\text{expression})=k\) in index form \(\text{expression}=a^{k}\) (or, if both sides are logs to the same base, equate the arguments).
Solve the resulting equation for \(x\) — often linear or quadratic.
Check the domain. Substitute each solution back; reject any that make the argument of a log zero or negative.

For an exponential equation take \(\ln\) of the whole side, use \(\ln a^{x}=x\ln a\) to drop the exponent, then solve the linear equation. For \(e^{2x-1}=7\): \(2x-1=\ln 7\), so \(x=\dfrac{1+\ln 7}{2}\). Leave exact answers in \(\ln\) or \(e^{k}\) form; use technology for a decimal.

Example 1 — Exponential, base \(e\)

Solve \(e^{2x-1}=7\), giving the exact value and a \(2\) d.p. value.

Solution

Take \(\ln\) of both sides so the exponent comes down.

\(e^{2x-1}\)	\(=\)	\(7\)
\(2x-1\)	\(=\)	\(\ln 7\)
\(x\)	\(=\)	\(\dfrac{1+\ln 7}{2}\approx 1.47\)

x = \frac{1 + \ln 7}{2}

Example 2 — Exponential, other base

Solve \(5^{x}=40\), giving the exact value.

Solution

Take \(\ln\); use \(\ln 5^{x}=x\ln 5\), then divide.

\(x\ln 5\)	\(=\)	\(\ln 40\)
\(x\)	\(=\)	\(\dfrac{\ln 40}{\ln 5}=\log_{5}40\)
	\(\approx\)	\(2.29\)

x = \frac{\ln 40}{\ln 5}

Example 3 — Logarithmic

Solve (a) \(\ln(x+2)=1\); (b) \(\log_{2}x=5\).

Solution

(a) Rewrite in index form (base \(e\)).

\(x+2\)	\(=\)	\(e^{1}\)
\(x\)	\(=\)	\(e-2\approx 0.72\)

(b) \(\log_{2}x=5\Rightarrow x=2^{5}=32\).

x = e - 2

Example 4 — Log laws + domain

Solve \(\log_{10}x+\log_{10}(x-3)=1\).

Solution

Combine with the product law, then rewrite in index form.

\(\log_{10}\!\big(x(x-3)\big)\)	\(=\)	\(1\)
\(x(x-3)\)	\(=\)	\(10\)
\(x^{2}-3x-10\)	\(=\)	\(0\)
\((x-5)(x+2)\)	\(=\)	\(0\)

\(x=5\) or \(x=-2\); the domain needs \(x>3\), so reject \(x=-2\). \(x=5\).

x = 5

Common pitfalls

Forgetting to check the domain. A log needs a positive argument. \(\log x+\log(x-3)=1\) gives \(x=5\) or \(x=-2\), but \(x=-2\) makes \(\log(-2)\) undefined, so it is rejected. Always substitute back.

Splitting the log of a sum. \(\log(a+b)\neq\log a+\log b\). The sum law is \(\log a+\log b=\log(ab)\); you can only combine or split a log across a product or quotient, never a sum.

Mixing bases. \(\ln\) is base \(e\) and \(\log\) (no base written) is base \(10\). Keep to one base throughout, and read off \(\log_r k=\dfrac{\ln k}{\ln r}\) with the change-of-base rule when the base is not \(e\) or \(10\).

Dropping the exponent too soon. For \(e^{2x-1}=7\), take \(\ln\) of the whole side to get \(2x-1=\ln 7\), then solve for \(x\). Writing \(x=\ln 7\) ignores the \(2x-1\).

Frequently asked questions

How do you solve an exponential equation like e^x = 7?

Take \(\ln\) of both sides (the inverse of \(e^x\)): \(e^x=7\) gives \(x=\ln 7\approx 1.95\).

How do you solve 5^x = 40 (a base other than e)?

Take \(\ln\) and use \(\ln 5^x=x\ln 5\): \(x=\dfrac{\ln 40}{\ln 5}=\log_5 40\approx 2.29\).

How do you solve a logarithmic equation like log_2 x = 5?

Rewrite in index form: \(\log_2 x=5\Rightarrow x=2^{5}=32\). Likewise \(\ln x=3\Rightarrow x=e^{3}\).

Why do you check the domain when solving log equations?

A log is defined only for a positive argument, so reject any solution that makes an argument \(\le 0\). \(\log x+\log(x-3)=1\Rightarrow x=5\) (rejecting \(x=-2\)).

How do you solve e^(2x-1) = 7 with a linear exponent?

Take \(\ln\) of the whole side: \(2x-1=\ln 7\), so \(x=\dfrac{1+\ln 7}{2}\approx 1.47\).

How do you solve a quadratic in disguise like 2^(2x) - 5(2^x) + 4 = 0?

Let \(y=2^x\); then \(y^2-5y+4=0\Rightarrow(y-1)(y-4)=0\), so \(2^x=1\Rightarrow x=0\) and \(2^x=4\Rightarrow x=2\).

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