Solve log(x) + log(x − 3) = 1
Logarithmic equation, worked out line by line the way a teacher would write it.
Answer
| Solution | x = 5 |
Step-by-step solution
15 steps-
1 Given\log_{10}{\left(x \right)} + \log_{10}{\left(x - 3 \right)} = 1
Solve for x.
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2 Domain: the argument of every logarithm must be positive
A logarithm is defined only for positive numbers, so every log argument must be greater than 0. Any answer that breaks this is rejected at the end.
x - 3 > 0,\quad x > 0\quad\Longrightarrow\quad x > 3 -
3 Combine the logarithms: log a + log b = log(ab), log a − log b = log(a/b), k·log a = log(aᵏ)
These rules turn several logarithms into one: log a + log b = log(ab), log a − log b = log(a/b) and k·log a = log(aᵏ). A single logarithm can then be removed by writing it in exponential form.
\log_{10}{\left(x \left(x - 3\right) \right)} = 1 -
4 Rewrite in exponential form: log_10(g) = d means g = 10^d
A logarithm answers the question “which power gives this number?”: log_b(g) = d means b^d = g. Rewriting it that way removes the logarithm.
x \left(x - 3\right) = 10 -
5 Expand the brackets and collect like terms
Multiply out each bracket with the distributive property, a(b + c) = ab + ac, then combine like terms: terms with the same power of the variable, such as 3x and −x, add up to 2x.
x^{2} - 3 x = 10 -
6 Move every term to the left side so the right side is 0
Factoring and the quadratic formula both work on an equation of the form … = 0. Subtracting the right side from both sides gets there without changing the solutions.
x^{2} - 3 x - 10 = 0 -
7 This is a quadratic in standard form ax² + bx + c = 0
Every quadratic equation can be arranged as ax² + bx + c = 0. Reading off a = 1, b = -3 and c = -10, signs included, shows which method fits: factoring, taking a square root or the quadratic formula.
a = 1,\quad b = -3,\quad c = -10 -
8 Factor the trinomial: find two numbers whose product is c = -10 and whose sum is b = -3
The goal is to write x² + bx + c as (x + p)(x + q). Multiplying that out gives x² + (p + q)x + p·q, so p and q must multiply to c = -10 and add up to b = -3.
List the pairs of numbers whose product is -10, with their signs, and pick the pair whose sum is -3.
(-5) \cdot 2 = -10,\qquad (-5) + 2 = -3 -
9 Write the factored form\left(x - 5\right) \left(x + 2\right) = 0
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10 Zero product property: a product is 0 only when one of its factors is 0
0 is the only number with this property: if a·b = 0, then a = 0 or b = 0. That is why the equation was first rearranged to … = 0 and factored: now each factor can be set to 0 on its own, giving a simpler equation for each.
x - 5 = 0\quad \text{or} \quad x + 2 = 0 -
11 Add 5 to both sides
An equation stays true as long as you do the same thing to both sides, like adding the same weight to both pans of a balance. Adding or subtracting the same amount on both sides moves a term to the other side, where it appears with the opposite sign.
x = 5Move the constant terms to the right.
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12 Subtract 2 from both sidesx = -2
Move the constant terms to the right.
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13 Reject x = -2: it makes a logarithm argument non-positive (extraneous)
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14 Remaining solutionx = 5
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15 Check\begin{aligned}x = 5:\quad \log_{10}{\left(2 \right)} + \log_{10}{\left(5 \right)} = 1\quad\checkmark\end{aligned}
Substituting each solution back makes both sides equal.
Check by substitution
| x | Left side | Right side | |
|---|---|---|---|
| 5 | log_10(2) + log_10(5) | 1 | ✓ |
Change a number or try your own problem: the solver works it out the same way, with a graph and hint mode.