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Math II

Exponential and logarithmic functions: practice problems

Exponents and logarithms are two views of the same relation: ap=Miffp=logaMa^p = M iff p = log_a M. Build fluency with the rules and with solving equations and inequalities.

Basic, Standard: Grade 11 · Term 2 / Advanced: Grade 12 · Exam prep

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Exponent laws and roots

Exponent laws (a>0a > 0)
apaq=ap+q,(ap)q=apq,(ab)p=apbp,amn=amn,a−p=1apa^pa^q = a^{p+q},\qquad (a^p)^q = a^{pq},\qquad (ab)^p = a^pb^p,\qquad a^{\frac{m}{n}} = \sqrt[n]{a^m},\qquad a^{-p} = \frac{1}{a^p}

Write everything as powers of the same base. Example: 43×163=223×243=22=4\sqrt[3]{4} \times \sqrt[3]{16} = 2^{\frac{2}{3}} \times 2^{\frac{4}{3}} = 2^2 = 4.

To compare numbers, use a common base: if the base is greater than 11, a larger exponent gives a larger number (the reverse if the base is less than 11).

Logarithm rules and change of base

Rules (a>0a > 0, a≠1a \neq 1, M>0M > 0, N>0N > 0)
log⁡aMN=log⁡aM+log⁡aN,log⁡aMN=log⁡aM−log⁡aN,log⁡aMp=plog⁡aM\log_a MN = \log_a M + \log_a N,\qquad \log_a \frac{M}{N} = \log_a M - \log_a N,\qquad \log_a M^p = p\log_a M
log⁡ab=log⁡cblog⁡ca(change of base)\log_a b = \frac{\log_c b}{\log_c a}\quad(\text{change of base})

Example: log⁡23⋅log⁡38=log⁡3log⁡2⋅3log⁡2log⁡3=3\log_2 3 \cdot \log_3 8 = \frac{\log 3}{\log 2} \cdot \frac{3\log 2}{\log 3} = 3.

Equations and inequalities

For exponential equations, write both sides as powers of the same base and compare exponents. For 4x−6⋅2x+8=04^x - 6 \cdot 2^x + 8 = 0, substitute t=2xt = 2^x (t>0t > 0).

For logarithmic equations and inequalities, first require the arguments to be positive. Example: log⁡2x+log⁡2(x−2)=3\log_2 x + \log_2(x - 2) = 3 needs x>2x > 2; then x(x−2)=8x(x - 2) = 8 gives x=4x = 4 (x=−2x = -2 is rejected).

Watch out

If 0<a<10 < a < 1, inequalities reverse: ap<aq  ⟺  p>qa^p < a^q \iff p > q and log⁡aM<log⁡aN  ⟺  M>N\log_a M < \log_a N \iff M > N.

Common logarithms and digits

A positive integer NN has nn digits exactly when 10n−1≤N<10n10^{n-1} \leq N < 10^n, i.e. n−1≤log⁡10N<nn - 1 \leq \log_{10} N < n.

Example: with log⁡102=0.3010\log_{10} 2 = 0.3010, log⁡10250=15.05\log_{10} 2^{50} = 15.05, so 2502^{50} has 1616 digits. Since the fractional part 0.050.05 is less than log⁡102\log_{10} 2, the leading digit is 11.

For 0<N<10 < N < 1: if −n≤log⁡10N<−(n−1)-n \leq \log_{10} N < -(n - 1), the first non-zero digit is at the nn-th decimal place.