// Numbas version: finer_feedback_settings {"name": "Solving log equations using exponentials", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "Solving log equations using exponentials", "advice": "", "variable_groups": [{"variables": [], "name": "Unnamed group"}], "ungrouped_variables": ["power", "d", "f", "g", "b", "a", "c", "ans"], "functions": {}, "extensions": [], "variablesTest": {"condition": "", "maxRuns": 100}, "type": "question", "statement": "", "preamble": {"js": "", "css": ""}, "tags": ["exp", "exponential", "exponentials", "logarithm", "logarithms", "logs", "solving", "Solving equations", "solving equations"], "parts": [{"showFeedbackIcon": true, "marks": 0, "scripts": {}, "type": "gapfill", "showCorrectAnswer": true, "prompt": "

Solve the following equation for $x$

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$\\displaystyle{\\var{d}\\log_\\var{b}(\\simplify{{a}x+{c}})+\\var{f}=\\var{g}   }.$

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$x=$ [[0]]

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Note: You can use the symbol ^ to signify powers, and / to signify division. Please ensure you use brackets correctly.

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We start solving the equation one operation at a time by doing the inverse to both sides, when we get to undoing the log we recall the definition of $\\log_b$, write the equation in index form and continue solving.

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Recall: The definition of $\\log_b$ says $\\log_b(a)=c$ is equivalent to $b^c=a$.

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$\\displaystyle{\\var{d}\\log_\\var{b}(\\simplify{{a}x+{c}})+\\var{f}  }$$=$$\\var{g}$
$\\displaystyle{\\var{d}\\log_\\var{b}(\\simplify{{a}x+{c}})  }$$=$$\\var{g-f}$(subtract $\\var{f}$ from both sides) 
$\\displaystyle{\\log_\\var{b}(\\simplify{{a}x+{c}})  }$$=$$\\var{power}$(divide both sides by $\\var{d}$)
$\\simplify[basic]{{b}^{power}}$$=$$\\simplify{{a}x+{c}}$(using the definition of $\\log_\\var{b}$)
$\\simplify[basic,unitpower]{{b}^{power}-{c}}$$=$ $\\var{a}x$(subtract $\\var{c}$ from both sides)
 
$\\displaystyle{\\simplify[basic,fractionnumbers,unitpower]{({b}^{power}-{c})/{a}}}$$=$$x$(divide both sides by $\\var{a}$)
 
$x$$=$$\\displaystyle{\\simplify[basic,fractionnumbers,unitpower]{{({b}^{power}-{c})/{a}}}}$
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