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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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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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