Loading the lesson
Loading the lesson
Search AllRounder.ai
Search the courses, subjects, tracks, games and features, or jump straight to a page.
Search the courses, subjects, tracks, games and features, or jump straight to a page.
Learn it your way
A short class with Sarah and her students. The other ways are under More.
Sign up free to listen. Transcripts are free to read.
Here is Example 1: log₂ x = 5. Before we solve it, how would you say this equation in words?
Log base 2 of x equals 5?
Good. Now say it as a question about powers: the power of 2 that gives x is 5.
So 2 to the power 5 equals x. That means x is 32.
Exactly. A logarithm is an exponent, so we rewrite log₂ x = 5 as 2⁵ = x.
And we can check it, because 2⁵ really is 32, so log₂ 32 is 5.
Suppose someone forgets the little 2 and writes log x = 5. What is the base now?
If no base is written, it means base 10.
Right. So what does log x = 5 give?
10⁵ = x, so x is 100 000. That is nothing like 32!
That is why we always read the base first. The base is the number that gets raised to the power.
So the pattern is always: base, raised to the answer of the log, equals the number inside.
Try log₅ x = −2.
x = 5⁻², which is 1 over 25. So x is positive even though the log is negative.
Good. Now a twist: logₓ 49 = 2. Here the base is the unknown.
Then x² = 49, so x is 7 or −7.
Can a base be negative?
No, a base must be positive and not 1, so only x = 7 works.
Excellent. Converting first and checking the conditions afterwards is the habit that keeps you safe.
log_b x = c means b^c = x, so the value of a logarithm is a power of the base.
rewriting log₂ x = 5 as 2⁵ = x turns the equation into arithmetic, giving x = 32.
log x with no base means base 10, so log x = 5 gives 100 000, while log₂ x = 5 gives 32.
the argument must be positive and the base must be positive and not equal to 1, which can rule out a solution.
substitute the answer back into the original equation to confirm it.
Solve log₂ x = 5: rewrite as x = 2⁵, so x = 32; check: 2⁵ = 32, so log₂ 32 = 5.
Solve log₃ x = 4: rewrite as x = 3⁴, so x = 81; check: 3⁴ = 81, so log₃ 81 = 4.
Solve log₅ x = −2: rewrite as x = 5⁻² = 1/25; the logarithm is negative, but x is still positive.
Solve logₓ 49 = 2: rewrite as x² = 49, so x = 7 or x = −7; reject −7 because a base must be positive, so x = 7.
The base goes up, the log's the power, the number's what you get; convert it first, then check it last, and you'll be right, I bet.
Base 2 is a little machine that doubles. Asked "log₂ x = 5", it simply runs five times: 2, 4, 8, 16, 32, and hands you x = 32.
"Base to the Log equals the Number" helps you write b^c = x from log_b x = c.
B.L.N. - Base, Log value, Number: the base raised to the log value gives the number.
Term
What does log₂ x = 5 mean in exponential form?
Click to flip
Term
What base is meant when log x has no base written?
Click to flip
Term
What must be true of the argument of a logarithm?
Click to flip
Term
What must be true of the base of a logarithm?
Click to flip
Term
How do you check a solution of a logarithmic equation?
Click to flip
The exponent to which a base must be raised to give a number; log_b x = c means b^c = x.
The number that is raised to a power in a logarithm; it must be positive and not equal to 1.
The number inside a logarithm, such as x in log₂ x; it must be positive.
A statement written as a power, such as 2⁵ = 32, equivalent to the logarithmic statement log₂ 32 = 5.
A logarithm with base 10, written log x when no base is shown.
Two functions that undo each other, such as y = 2^x and y = log₂ x.
Putting a solution back into the original equation to confirm that it works.
Example 1 solves the logarithmic equation log₂ x = 5 by rewriting it in exponential form, which gives x = 2⁵ = 32. The same one-step conversion is the starting point for every logarithmic equation in this chapter.
A logarithm is an exponent: log₂ x = 5 says that 5 is the power you must raise 2 to in order to get x. Rewriting the statement as 2⁵ = x turns a logarithmic equation into a simple calculation, so x = 32. This lesson works through Example 1 step by step, checks the answer, and then practises the same conversion with other bases, negative and fractional exponents, and an unknown base. It also shows why the base must always be read carefully: log₂ x = 5 and log x = 5 have very different answers.
Example 1 of "Solving Logarithmic Equations" is the simplest kind of logarithmic equation: the unknown is the number inside the logarithm, and the base and the value of the logarithm are both known.
The key fact. For a base b with b > 0 and b ≠ 1, and a number x > 0:
log_b x = c means exactly the same as b^c = x.
The logarithm is the exponent. Reading the logarithmic statement aloud as "the power of b that gives x is c" makes the conversion feel natural.
Example 1: Solve log₂ x = 5.
Watch the base. When no base is written, log x means the common logarithm, base 10. So log x = 5 would give x = 10⁵ = 100 000, not 32. The small subscript 2 in log₂ x changes the answer completely, so always copy the base carefully.
The same method with other numbers.
When the base is the unknown. The conversion still works. For logₓ 49 = 2, rewrite as x² = 49, so x = 7 or x = −7. A base must be positive and not equal to 1, so x = −7 is rejected and the only solution is x = 7.
Why it works. The functions y = 2^x and y = log₂ x are inverses: each undoes the other. Rewriting log₂ x = 5 as x = 2⁵ is simply applying the exponential function to both sides to undo the logarithm.
Points to remember.
Examples 2 and 3 of this section build on this skill: Example 2 has an expression such as x + 1 inside the logarithm, and Example 3 first combines two logarithms with the product rule. In both, the key step is the same conversion you practise here.