// Numbas version: exam_results_page_options {"name": "Functions: Inverse Functions 4", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "Functions: Inverse Functions 4", "tags": [], "metadata": {"description": "

Finding the inverse of a function of the form $f(x)=e^{mx+c}$.

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If $f(x)=e^\\simplify{{m}x+{c}}$, find the inverse function, $f^{-1}(x)$.

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To find $f^{-1}x$, it can help to first set $f(x)$ to a different variable, which we will call $y$:

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\\[ y = f(x) = e^\\simplify{{m}x+{c}}\\]

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Since the function $f(x)$ takes us from $x$ to $y$, the inverse function $f^{-1}$ will take us from $y$ to $x$. So to obtain $f^{-1}$, we want to rearrange $y=e^\\simplify{{m}x+{c}}$ so that it is $x$ as a function of $y$:

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\\[ \\begin{split} y &\\,= e^\\simplify{{m}x+{c}} \\\\ \\ln(y) &\\,= \\simplify{{m}x+{c}} \\\\ \\simplify{ln(y)-{c}} &\\,= \\simplify{{m}x} \\\\ \\implies x &\\,= \\simplify{(ln(y)-{c})/{m}}. \\end{split} \\]

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Hence, $f^{-1}(y) =\\simplify{(ln(y)-{c})/{m}}$, and therefore \\[ f^{-1}(x) =\\simplify{(ln(x)-{c})/{m}}.\\]

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(Note: The inverse is valid for all values of $x>0$.)

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$f^{-1}(x)=$[[0]]

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