To answer the question above, directly substitute the given price which is $100 to all the x's of the given equation,
C = 0.765x + 0.06 (0.765x)
C = (0.765) ($100) + 0.06 (0.765) ($100) = 81.09
Thus, Chris would have to pay only $81.09.

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3 years ago

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Zeke deposited $300 into a savings account. The interest rate on the account is 3%, compounded continuously. To determine how lo

Vedmedyk [2.9K]

$\bf \begin{array}{llll} \textit{logarithm of factors} \\\\ \log_a(xy)\implies \log_a(x)+\log_a(y) \end{array} ~\hspace{4em} \begin{array}{llll} \textit{Logarithm of rationals} \\\\ \log_a\left( \frac{x}{y}\right)\implies \log_a(x)-\log_a(y) \end{array}$

$\bf \begin{array}{llll} \textit{Logarithm of exponentials} \\\\ \log_a\left( x^b \right)\implies b\cdot \log_a(x) \end{array} ~\hspace{7em} \begin{array}{llll} \textit{Logarithm Cancellation Rules} \\\\ \stackrel{\textit{let's use this one}}{log_a a^x = x}\qquad \quad a^{log_a x}=x \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ 900=300e^{0.03t}\implies \log_e(900)=\log_e(300e^{0.03t})$

$\bf \log_e(900)=\log_e(300)+\log_e(e^{0.03t})\implies \ln(900)=\ln(300)+0.03t\cdot \ln(e) \\\\\\ \ln(900)-\ln(300)=0.03t\implies \ln\left( \cfrac{900}{300} \right)=0.03t\implies \ln(3)=0.03t$

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3 years ago