// Numbas version: exam_results_page_options {"name": "Logarithms: Solving Logarithmic Equations 7 - Sum of Logarithms", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "Logarithms: Solving Logarithmic Equations 7 - Sum of Logarithms", "tags": [], "metadata": {"description": "

Solving $\\log(y)+\\log(x)=\\frac{1}{n}\\log(ay^n)$ for $x$, where $a$ and $n$ are positive integers.

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Solve for $x$:

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\\[ \\log(y)+\\log(x)=\\frac{1}{\\var{n}}\\log(\\var{a^n}y^\\var{n}).\\]

", "advice": "

To solve for $x$, we want to make use of the following logarithm rules:

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Therefore, 

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\\[ \\begin{split} &\\log(y)+&\\log(x) &\\,= \\frac{1}{\\var{n}} \\log(\\var{a^n}y^\\var{n}) \\\\ &\\implies &\\log(yx) &\\,= \\log\\left((\\var{a^n}y^\\var{n})^\\frac{1}{\\var{n}})\\right) \\\\ &\\implies &\\log(yx) &\\,= \\log(\\var{a}y) \\end{split} \\]

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Recall that if $\\log(a) = \\log(b)$, then $a=b$.

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Hence,

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\\[ \\begin{split}  \\log(yx) &\\,= \\log(\\var{a}y) \\\\ \\implies \\quad yx &\\,= \\var{a}y \\\\ \\implies \\quad x &\\,= \\var{a}. \\end{split} \\]

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$x=$[[0]]

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