Exponential equations and models

Each year retains a fraction 1 − p/100 of that year starting amount. After n years the value is S × (1 − p/100)ⁿ. Apply each decrease to the already reduced amount.

A percentage-increase multiplier must include both the entire starting value and its increase.

1. The starting value is 100%, and the increase adds another p%. The new value is therefore (100 + p)% of the starting value.
2. Convert this percentage into a decimal by dividing by 100.
3. The multiplier is q = (100 + p) / 100 = 1 + p / 100.

Multiplying by p / 100 alone gives only the increase, not the whole new value. Check that an increase multiplier is greater than 1.

The exponential aˣ increases for a > 1 and decreases for 0 < a < 1. Comparing powers of an increasing base preserves the exponent inequality; a decreasing base reverses it. For 1 < a < b and x > 0, aˣ < bˣ.

For a > 0 and a ≠ 1, equal powers of a have equal exponents. From a^(x + b) = aᶜ obtain x + b = c, hence x = c − b. For aˣ = v, x = logₐ v when v > 0.

When a value grows by the same percentage each year, each increase is calculated from the already increased value. This is compound growth.

1. Convert the annual growth rate p percent into a multiplier: q = 1 + p / 100.
2. Multiply the initial value S by q once for each year.
3. After n years, S_n = S · (1 + p / 100)ⁿ.
4. Keep intermediate calculations precise. If the answer is shown to the nearest hundredth, round to two decimal places only at the end.

Do not replace the power by simply multiplying the percentage by the number of years: that would apply all increases to the initial value only, rather than compound them.

Replace the exponential expression by a positive variable. Solve the quadratic, recover x, then add the solutions.

For a > 0 and a ≠ 1, equal powers of a have equal exponents. From a^(x + b) = aᶜ obtain x + b = c, hence x = c − b. For aˣ = v, x = logₐ v when v > 0.

For a > 0 and a ≠ 1, equal powers of a have equal exponents. From a^(x + b) = aᶜ obtain x + b = c, hence x = c − b. For aˣ = v, x = logₐ v when v > 0.

For a > 0 and a ≠ 1, equal powers of a have equal exponents. From a^(x + b) = aᶜ obtain x + b = c, hence x = c − b. For aˣ = v, x = logₐ v when v > 0.

The exponential aˣ increases for a > 1 and decreases for 0 < a < 1. Comparing powers of an increasing base preserves the exponent inequality; a decreasing base reverses it. For 1 < a < b and x > 0, aˣ < bˣ.