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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.
\nRecall: The definition of $\\log_b$ says $\\log_b(a)=c$ is equivalent to $b^c=a$.
\n$\\displaystyle{\\var{d}\\log_\\var{b}(\\simplify{{a}x+{c}})+\\var{f} }$ | \n$=$ | \n$\\var{g}$ | \n\n |
$\\displaystyle{\\var{d}\\log_\\var{b}(\\simplify{{a}x+{c}}) }$ | \n$=$ | \n$\\var{g-f}$ | \n(subtract $\\var{f}$ from both sides) | \n
$\\displaystyle{\\log_\\var{b}(\\simplify{{a}x+{c}}) }$ | \n$=$ | \n$\\var{power}$ | \n(divide both sides by $\\var{d}$) | \n
$\\simplify[basic]{{b}^{power}}$ | \n$=$ | \n$\\simplify{{a}x+{c}}$ | \n(using the definition of $\\log_\\var{b}$) | \n
$\\simplify[basic,unitpower]{{b}^{power}-{c}}$ | \n$=$ | \n$\\var{a}x$ | \n(subtract $\\var{c}$ from both sides) | \n
\n | \n | \n | \n |
$\\displaystyle{\\simplify[basic,fractionnumbers,unitpower]{({b}^{power}-{c})/{a}}}$ | \n$=$ | \n$x$ | \n(divide both sides by $\\var{a}$) | \n
\n | \n | \n | \n |
$x$ | \n$=$ | \n$\\displaystyle{\\simplify[basic,fractionnumbers,unitpower]{{({b}^{power}-{c})/{a}}}}$ | \n\n |
Solve the following equation for $x$
\n$\\displaystyle{\\var{d}\\log_\\var{b}(\\simplify{{a}x+{c}})+\\var{f}=\\var{g} }.$
\n\n$x=$ [[0]]
\nNote: You can use the symbol ^ to signify powers, and / to signify division. Please ensure you use brackets correctly.
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