Log Questions (1 Viewer)

Zak Ambrose · May 18, 2009

1. if log A = bt + log P, express A in terms of the other symbols.

2. log 2 in base "a" = log 16 in base "b", show that b = a^4

3.find the values of x for which
(log x)(log x^2) + log x^3 - 5 = 0

gurmies · May 18, 2009

$Question \,\, 1 \\\\ \log A = bt + \log P \\\\ \log A - \log P = bt \\\\ \log \frac{A}{P} = bt \\\\ \frac{A}{P} = 10^{bt} \\\\ A = P.10^{bt}$

$Question \,\, 3 \\\\ (\log x)(\log x^{2}) + \log x^{3} - 5 = 0 \\\\ 2(\log x)^{2} + 3\log x - 5 = 0 \\\\ (2\log x + 5)(\log x - 1)=0 \\\\ \log x = -\frac{5}{2} \,\, or/ \,\, \log x = 1 \\\\ x = 10^{\frac{-5}{2}} \,\, or/ \,\, x = 10$

Ambiguous as i'm not sure of the base xD

Zak Ambrose · May 20, 2009

base is 10.
thanks for the help!!!

Zak Ambrose · May 26, 2009

i have some more log/eponential questions

1. 0.3xe^(-1.1x) = 0.06, find x
2. find the min value of y = e^x + 4e^-x
3. graph y = (e^-x)(cosx) in the domain 0< x < pi
4. find the area enclosed by y = e^x, the y-axis and the line y = e^2

Drongoski · May 27, 2009

Q4.

$\100dpi \texttt{} \ \ y\ =\ e^{x}\ \ \Rightarrow \ \ x\ =\ ln\ y\\ \\ soln1:\ \ \ A\ =\ \int_{1}^{e^{2}}x\ dy\ \ =\ \ \int_{1}^{e^{2}}lny\ dy\\ \\ \ \ =\ \ \left [ y\ \cdot lny \right ]^{e^{2}}_{1}\ -\ \int_{1}^{e^{2}}y\cdot \frac{1}{y}\ dy\ \ \ (integration\ by\ parts)\\ \\ \ \ =\ \ \left [ y\cdot lny\ -\ y \right ]^{e^{2}}_{1}\ \ =\ \ (e^{2}ln\ e^{2}-e^{2})\ -\ (1\cdot ln\ 1-1)\\ \\ \ \ =\ \ (2e^{2}-e^{2})-(0-1)\ \ =\ \ e^{2}+1\\ \\ \\ Soln2: (if\ you\ do \ only\ 2u\ or\ 3u\ Maths,\ u\ don't\ do\ integn\ by\ parts)\\ \\ \ A\ =\ Area\ of\ rectangle\ -\ \int_{0}^{2}y\ dx\ \ =\ \ 2e^{2} \ -\ \int_{0}^{2}e^{x}\ dx\\ \\ \ \ =\ \ 2e^{2}\ -\ \left [ e^{x} \right ]^{2}_{0}\ \ =\ \ 2e^{2}\ -\ e^{2}+1\ \ =\ \ e^{2}+1$

Zak Ambrose · May 27, 2009

Drongoski said:
$\100dpi \texttt{} \ \ y\ =\ e^{x}\ \ \Rightarrow \ \ x\ =\ ln\ y\\ \\ [color="DarkOrchid"]soln1[/color]:\ \ \ A\ =\ \int_{1}^{e^{2}}x\ dy\ \ =\ \ \int_{1}^{e^{2}}lny\ dy\\ \\ \ \ =\ \ \left [ y\ \cdot lny \right ]^{e^{2}}_{1}\ -\ \int_{1}^{e^{2}}y\cdot \frac{1}{y}\ dy\ \ \ (integration\ by\ parts)\\ \\ \ \ =\ \ \left [ y\cdot lny\ -\ y \right ]^{e^{2}}_{1}\ \ =\ \ (e^{2}ln\ e^{2}-e^{2})\ -\ (1\cdot ln\ 1-1)\\ \\ \ \ =\ \ (2e^{2}-e^{2})-(0-1)\ \ =\ \ e^{2}+1\\ \\ \\ [color="DarkOrchid"]Soln2[/color]: (if\ you\ do \ only\ 2u\ Maths,\ u\ don't\ do\ integn\ by\ parts)\\ \\ \ A\ =\ Area\ of\ rectangle\ -\ \int_{0}^{2}y\ dx\ \ =\ \ 2e^{2} \ -\ \int_{0}^{2}e^{x}\ dx\\ \\ \ \ =\ \ 2e^{2}\ -\ \left [ e^{x} \right ]^{2}_{0}\ \ =\ \ 2e^{2}\ -\ e^{2}+1\ \ =\ \ e^{2}+1$

which one?

ah, dw bout Q1.

Drongoski · May 27, 2009

Zak Ambrose said:
which one?

ah, dw bout Q1.

Q4. The comment appeared when I tried to highlight Soln1 and Soln2 in colour.

Log Questions (1 Viewer)

Zak Ambrose

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Zak Ambrose

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Zak Ambrose

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Drongoski

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