6.2 Exponential functions
Definition 6.2. For a positive real \(a\neq 1\), the exponential function with base \(a\) is \[f(x)=a^x.\]
The base must be positive (otherwise \(a^{1/2}\) has no real value) and not \(1\) (since \(1^x\) is the constant \(1\), which is not interesting).
Note 6.3. Every exponential graph passes through \((0,1)\), because \(a^0=1\) whatever \(a\) is. The domain is all of \(\mathbb {R}\) and the range is \((0,\infty )\) — an exponential is never zero or negative, however far left you go. The \(x\)-axis is a horizontal asymptote it approaches but never reaches.
6.2.1 Exponential equations
If both sides can be written with the same base, the indices can be equated.
Solution. (a) Write \(256\) as a power of \(4\): \(4^1=4\), \(4^2=16\), \(4^3=64\), \(4^4=256\). \[4^x=4^4\quad \implies \quad x=4.\]
(b) First turn the decimal into a fraction: \(0.125=\frac {1}{8}\). Then \[\frac {1}{8}=\frac {1}{2^3}=2^{-3},\] so \[2^x=2^{-3}\quad \implies \quad x=-3.\]
(c) \(27=3^3\), so \[3^{2x-1}=3^3\quad \implies \quad 2x-1=3\quad \implies \quad 2x=4\quad \implies \quad x=2.\]
Check. \(3^{2(2)-1}=3^3=27\). \(\relax \amscheckmark \)
Solution. This looks unlike the others, but notice \(2^{2x}=\left (2^x\right )^2\). Substituting \(u=2^x\) turns it into a quadratic: \[u^2-6u+8=0\quad \implies \quad (u-2)(u-4)=0\quad \implies \quad u=2 \text { or } u=4.\]
Now go back to \(x\) — the substitution must always be undone. \[2^x=2\ \implies \ x=1,\qquad 2^x=4=2^2\ \implies \ x=2.\] \[\therefore \quad x=1 \text { or } x=2.\]
Check. At \(x=2\): \(2^4-6(2^2)+8=16-24+8=0\). \(\relax \amscheckmark \)
Note 6.6. Had a root come out negative or zero — say \(u=-3\) — it would have to be discarded, since \(2^x\) is never negative. Always check that each value of \(u\) is actually attainable before solving for \(x\).
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