Sometimes tested Papers 1 & 2

Exponentials and Logarithms: Graphs, Log Laws and Solving a^x = b

Exponential and logarithmic functions are inverses of one another: y = a^x reflects onto y = loga(x) in the line y = x. The marks come from the log laws (product, quotient, power) and no-calculator methods for solving a^x = b, including same-base equations and hidden quadratics such as 25^x - 6 × 5^x + 5 = 0.

Part of the TMUA syllabus — revision for the Test of Mathematics for University Admission (TMUA), the UAT-UK maths test used by Cambridge, Oxford, Imperial, UCL, LSE, Warwick and Durham. No calculator; multiple choice.

Key points

  • y = a^x with a > 1 is increasing, always positive, and passes through (0, 1).
    a0 = 1
  • For 0 < a < 1 the curve is decreasing and is the reflection of the a > 1 graph in the y-axis, still passing through (0, 1).
  • Every exponential graph has the x-axis as an asymptote: the range is y > 0 and the domain is all real x (a^x is never 0 or negative).
  • The logarithm is the inverse of the exponential: a^x = b is exactly the same statement as x = loga(b).
  • y = loga(x) is the reflection of y = a^x in the line y = x, so it is defined only for x > 0.
  • Log laws convert products into sums and powers into multiples (loga(x^k) = k × loga(x)).
  • To solve a^x = b, take logs of both sides and use the power rule: x = log(b)/log(a).
  • A disguised quadratic such as 25^x - 6 × 5^x + 5 = 0 becomes y2 - 6y + 5 = 0 under the substitution y = 5^x.

Diagram

xy-2024-224y = 2^xy = log_2 xy = x
y = a^x and y = loga(x) are reflections of each other in the line y = x (shown for a = 2).

Formulae

loga(xy) = loga(x) + loga(y)

Product rule: split or combine the log of a product into a sum of logs.

loga(x/y) = loga(x) - loga(y)

Quotient rule: turn the log of a fraction into a difference of logs.

loga(x^k) = k × loga(x)

Power rule: bring an exponent down as a multiplier. Essential for solving a^x = b.

loga(1) = 0 and loga(a) = 1

Special values that simplify expressions and let you eliminate multiple-choice options fast.

a^x = b ≤> x = loga(b)

Switch between exponential form and log form. This is the definition of a logarithm.

loga(x) = logb(x) / logb(a)

Change of base, for example to rewrite log2(5) using base-10 or natural logs on paper.

Definitions

Exponential function
A function y = a^x where the base a is a fixed positive constant and the exponent x is the variable. It gives growth if a > 1 and decay if 0 < a < 1.
a ≠ 1
Logarithm
loga(b) is the power to which the base a must be raised to give b, so loga(b) = x means the same as a^x = b.
with a > 0, a ≠ 1 and b > 0
Common and natural logs
log (also written lg) means log base 10, and ln means log base e, where e is about 2.718. Both obey the same laws and let you solve a^x = b by hand.
Asymptote
A line that a curve approaches but never reaches. For y = a^x the x-axis is a horizontal asymptote.
the line y = 0

Worked examples

1

Solve for x: log2(x) + log2(x - 2) = 3.

  1. 1

    Combine the two logs with the product rule:

    log2(x(x-2)) = 3.

  2. 2

    Rewrite in exponential form:

    x(x-2) = 23
  3. 3

    Simplify the right-hand side:

    x(x-2) = 8
  4. 4

    Expand and rearrange:

    x2 - 2x - 8 = 0.

  5. 5

    Factorise the quadratic:

    (x-4)(x+2) = 0
  6. 6

    Read off the roots:

    x = 4 or x = -2
  7. 7

    Reject the negative root because log2 needs a positive argument:

    x = 4

Answer: x = 4 (x = -2 is rejected because log2 is only defined for positive arguments).

2

Solve for x: 25^x - 6 × 5^x + 5 = 0.

  1. 1

    Write the first term as a square:

    25^x = (5^x)2
  2. 2

    Substitute y = 5^x:

    y2 - 6y + 5 = 0.

  3. 3

    Factorise the quadratic:

    (y-1)(y-5) = 0
  4. 4

    Solve for y:

    y = 1 or y = 5
  5. 5

    Back-substitute y = 5^x:

    5^x = 1 or 5^x = 5
  6. 6

    Solve each (both values are positive, so both are valid):

    x = 0 or x = 1

Answer: x = 0 or x = 1.

Common mistakes

  • ×Treating loga(x + y) as loga(x) + loga(y). The sum rule applies to a product: loga(xy) = loga(x) + loga(y), not to a sum.
  • ×Forgetting the domain: the log of zero or a negative number is undefined, so always reject any root that makes a log's argument ≤ 0.
  • ×Confusing log(x2), (log x)2 and 2 log x. Only log(x2) = 2 log x; the power must be inside the log for the power rule.
  • ×Assuming a^x can be zero or negative. For a > 0 the graph stays strictly above the x-axis (y > 0), so an equation like a^x = -2 has no solution.
  • ×Solving a^x = b by dividing instead of taking logs. The exponent comes down only via logs: x = log(b)/log(a), not b/a.

No-calculator tips

  • ✓Rewrite both sides with the same base whenever you can (8 = 23, 25 = 52, 9 = 32), then simply equate the exponents so no logarithms are needed.
  • ✓Memorise the special values a0 = 1, a1 = a, loga(1) = 0 and loga(a) = 1; they collapse many multiple-choice options instantly.
  • ✓For a^x = b where b is a power of a, read the answer straight off (for example 5^x = 125 gives x = 3).
  • ✓Spot a hidden quadratic when you see a2x (or 25^x, 4^x) sitting alongside a^x, and substitute y = a^x.
  • ✓Use the graph shape to eliminate options: an increasing exponential through (0, 1) forces x = 0 when the right-hand side is 1, and any candidate that makes a log argument negative can be discarded.

Test yourself

Original practice questions, no calculator. Work each out before revealing the answer.

Q1.Find the complete set of real solutions of 9^x - 4 × 3^x + 3 = 0.

  • A. x = 1 and x = 3
  • B. x = 0 and x = 1
  • C. x = 1 only
  • D. x = 0 only
  • E. no real solutions
Show answer

Answer: B — x = 0 and x = 1

Let y = 3^x (so 9^x = (3^x)2 = y2); then y2 - 4y + 3 = (y-1)(y-3) = 0 gives y = 1 or y = 3, i.e. 3^x = 1 ⇒ x = 0 and 3^x = 3 ⇒ x = 1. Option (a) is the trap of quoting the y-values as x.

Q2.Solve for real x: log3(x) + log3(x - 6) = 3.

  • A. x = 9 and x = -3
  • B. x = -3
  • C. x = 16.5
  • D. x = 9
  • E. x = 27
Show answer

Answer: D — x = 9

Combining gives log3(x(x-6)) = 3, so x2 - 6x = 27, giving (x-9)(x+3) = 0; x = -3 is rejected since log needs x > 6, leaving x = 9. Option (c) comes from wrongly adding the arguments (2x - 6 = 27).

Q3.Which single transformation maps the graph of y = 2^x onto the graph of y = log2(x)?

  • A. reflection in the x-axis
  • B. reflection in the y-axis
  • C. reflection in the line y = x
  • D. reflection in the line y = -x
Show answer

Answer: C — reflection in the line y = x

log2 is the inverse function of 2^x, and a function and its inverse are reflections of each other in the line y = x. Reflecting in an axis would instead give y = -2^x or y = 2^(-x), not a logarithm.

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