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a) 

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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 exponential we apply a log to both sides, we then use a log law and continue solving.

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$\\var{a}$$=$$\\simplify{{p}({b})^(n/{d})+{c}}$
$\\simplify{{a-c}}$$=$$\\simplify{{p}({b})^(n/{d})}$(subtract $\\var{c}$ from both sides) 
$\\var{frac}$$=$$\\simplify{{b}^(n/{d})}$(divide both sides by $\\var{p}$)
$\\log(\\var{frac})$$=$$\\log(\\var{b}^{\\frac{n}{\\var{d}}})$(take the log of both sides)
$=$ $\\frac{n}{\\var{d}}\\log(\\var{b})$(use a log law)
$\\displaystyle{\\frac{\\log(\\var{frac})}{\\log(\\var{b})}}$$=$$\\frac{n}{\\var{d}}$(divide both sides by $\\log(\\var{b})$)
$\\displaystyle{\\frac{\\var{d}\\log(\\var{frac})}{\\log(\\var{b})}}$$=$$n$(multiply both sides by $\\var{d}$)
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b)

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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 exponential we apply a log to both sides, we then use a log law and continue solving.

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${\\var{FV}}$$=$$\\displaystyle{\\simplify{{pay}((1+{int})^n-1)/{int}}}$
$\\simplify{{FV*int}}$$=$$\\displaystyle{\\simplify{{pay}((1+{int})^n-1)}}$(multiply both sides by $\\var{int}$) 
$\\displaystyle{\\simplify[simplifyFractions]{{FV*int}/{pay}}}$$=$$\\displaystyle{\\simplify{(1+{int})^n-1}}$(divide both sides by $\\var{pay}$)
$\\displaystyle{\\simplify[simplifyFractions]{{FV*int}/{pay}+1}}$$=$$\\displaystyle{\\simplify{(1+{int})^n}}$(add $1$ to both sides)
$\\displaystyle{\\simplify[simplifyFractions]{{FV*int+pay}/{pay}}}$$=$ $\\displaystyle{\\simplify{(1+{int})^n}}$(tidy up left hand side)
$\\displaystyle{\\log\\left(\\simplify[simplifyFractions]{{FV*int+pay}/{pay}}\\right)}$$=$$\\displaystyle{\\log\\left((\\var{1+int})^n\\right)}$(take the log of both sides)
$\\displaystyle{\\log\\left(\\simplify[simplifyFractions]{{FV*int+pay}/{pay}}\\right)}$$=$$\\displaystyle{n\\log(\\var{1+int})}$(use a log law)
$\\displaystyle{\\frac{\\log\\left(\\simplify[simplifyFractions]{{FV*int+pay}/{pay}}\\right)}{\\log(\\var{1+int})}}$$=$$n$(divide both sides by $\\log(\\var{1+int})$)
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Solve the following equation for $n$

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$\\begin{align*}\\simplify{{a}={p}({b})^(n/{d})+{c}}.\\end{align*}$

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$n=$ [[0]]

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Note: Typing $\\log(5)$ will input the value $\\log_{10}(5)$, whereas $\\log5$ will not work.
Note: Typing $\\ln(5)$ will input the value $\\log_e(5)$, whereas $\\ln5$ will not work.

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Solve the following equation for $n$

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$\\displaystyle{\\simplify{{FV}={pay}((1+{int})^n-1)/{int}}}.$

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$n=$ [[0]]

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Note: Typing $\\log(5)$ will input the value $\\log_{10}(5)$, whereas $\\log5$ will not work.
Note: Typing $\\ln(5)$ will input the value $\\log_e(5)$, whereas $\\ln5$ will not work.

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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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