For y = b^x with b > 0 and b ≠ 1: the domain is all real x, the range is y > 0, the y-intercept is (0, 1) and the x-axis (y = 0) is a horizontal asymptote. If b > 1 the graph increases (growth); if 0 < b < 1 it decreases (decay).

Transformations work as for other functions: y = b^x + c moves the graph and its asymptote up c units; y = a × b^x stretches it vertically; y = b^−x reflects it in the y-axis.

Worked example. y = 2^x + 3 has y-intercept 2⁰ + 3 = 4, asymptote y = 3 and range y > 3.

Same base: if both sides can be written as powers of the same number, equate the indices: b^m = bⁿ means m = n.

Different bases: take logarithms (ln or log) of both sides and use log(b^x) = x log b. So b^x = c gives x = [frac ln c/ln b]. Get the power on its own first.

Quadratic form: b^2x = (b^x)², so an equation like b^2x + pb^x + q = 0 becomes a quadratic in u = b^x. Remember that b^x > 0, so reject negative values of u.

Worked example. Solve 9^{x + 1} = 27^x. Powers of 3: 3^{2x + 2} = 3^3x, so 2x + 2 = 3x and x = 2.

A quantity that changes by the same percentage each period is modelled by y = a × b[sup t], where a is the starting amount. Growth of r% per period gives b = 1 + r100; decay of r% gives b = 1 − r100.

If a quantity doubles every d units of time, y = a × 2^t/d. If it has a half-life of h, y = a × (12)^t/h.

Worked example. A car bought for $30 000 loses 15% of its value each year. After 4 years V = 30 000 × 0.85⁴ = $15 660.19.