Begin with an exponent as a count of factors
In 2⁴ = 2 × 2 × 2 × 2 = 16, two is the base and four is the exponent. Multiplying powers with the same base joins their factors, giving a^m × a^n = a^(m + n). First read this with positive integers m and n by putting the two products in one row.
For a > 0, setting a⁰ = 1 and a^(−n) = 1/a^n preserves the exponent rule for zero and negative integers. A negative exponent does not mean a negative value. The power a^(1/2) is the positive square root of a. Moving from factor counts to roots, and then extending to real exponents, retains the rule that multiplying powers adds their exponents.
Read the same relationship from the exponent side
The statement log_a b = t means a^t = b: the t-th power of a is b. Here a is the base, b the argument, and t the exponent being found. Keep those three roles separate. A logarithm is neither the product of a and b nor b divided by a.
In the example 2^(x + 1) = 16, rewrite sixteen as 2⁴. Different real exponents of base two give different values, so x + 1 = 4 and x = 3. Four is the value of the entire exponent, not of x. Substituting x = 3 into the original equation gives 2⁴ = 16 as the final check.
Trace why the logarithm of a product is a sum
The product example specifies x > 0 and y > 0. Write u = log₁₀x and v = log₁₀y. In reverse, x = 10ᵘ and y = 10ᵛ, so xy = 10ᵘ × 10ᵛ = 10^(u + v). The power of ten giving xy is therefore u + v: log₁₀(xy) = log₁₀x + log₁₀y.
The derivation first restores the arguments to power form and then adds exponents of the same base. Multiplying the two logarithms does not follow that step. The argument inside the parentheses is the product xy. Replacing it with the sum x + y does not preserve this rule.
Put the original base in the denominator
For the change-of-base example, set t = log_a b, so a^t = b. With another base c, write r = log_c a and s = log_c b. Thus a = c^r and b = c^s. Substitution gives (c^r)^t = c^s, or c^(rt) = c^s. The one-to-one relationship between exponents and values for base c gives rt = s.
Because a ≠ 1, r = log_c a is nonzero and can be divided out. Thus t = s/r, giving log_a b = log_c b / log_c a. The argument b supplies the numerator; the original base a supplies the denominator. If the formula is hard to recall, starting from a^t = b recovers that order as well.
Keep the permitted values attached to the formula
A real logarithm requires a > 0, a ≠ 1 and b > 0. Base one gives 1^t = 1 for every exponent, so it cannot identify one exponent. Real powers of a positive base are positive, so zero and negative arguments are outside this logarithm. An argument of one is allowed: a⁰ = 1 gives log_a 1 = 0.
A base need not exceed one. For 0 < a < 1, increasing the exponent decreases the value, but the relationship remains one-to-one. In change of base, the new base must also satisfy c > 0 and c ≠ 1. Retain both the original-base condition a ≠ 1, which makes the denominator nonzero, and the conditions for the new base.
Return the three examples to power form
The examples below follow the order exponential equation, product rule, then change of base. Before opening an answer, write the corresponding power equation beside the logarithmic statement and circle the quantity being found. After reading, cover the explanation and reconstruct the exponent-addition step and the change-of-base step rt = s. Practise reversing the same relationship rather than adding new numbers.
If the unknown’s role or substitution is unclear, return to the prerequisite “Expressions: give the unknown a role”. The related guide “Graphs and coordinates: read what each axis means” offers a comparison with identifying inputs and outputs on axes. The practice preset selects upper-secondary, standard questions in “Numbers and calculation” and “Expressions and functions”, so it includes questions beyond these three examples.