Solving Exponential & Log Equations

When the variable is in the exponent, take the log of both sides and use the power rule.

For 3ˣ = 20:
log(3ˣ) = log 20 → x·log 3 = log 20 → x = log 20 / log 3 ≈ 2.727.

If both sides can be written with the same base, that is faster. 2ˣ = 32 is just 2ˣ = 2⁵, so x = 5.

When the variable is inside a log, condense to one log and then exponentiate.

log(x) + log(x − 3) = 1 → log(x(x−3)) = 1 → x(x−3) = 10 → x² − 3x − 10 = 0 → x = 5 or x = −2.

Then check. x = −2 makes log(−2) undefined, so only x = 5 survives.

Because exponential and logarithmic functions are one-to-one, applying either to both sides of an equation preserves the solution set on the appropriate domain. That injectivity also justifies the one-to-one property: b^u = b^v ⟹ u = v and log_b u = log_b v ⟹ u = v.

For exponential equations with non-matching bases, taking a logarithm of both sides and applying the power rule linearises the unknown. Either base works; ln is conventional for continuous models since ln(e^{kt}) = kt directly.

For logarithmic equations, condense to a single logarithm, then exponentiate. Because condensing with the product rule can enlarge the domain — log x + log(x−3) requires x > 3, while log(x² − 3x) permits x < 0 as well — extraneous solutions are systematically produced and must be discarded by checking against the original.

Applications include solving A = P(1 + r/n)^{nt} for t, and half-life problems A = A₀(1/2)^{t/h}, both of which require logarithms precisely because the unknown sits in the exponent.