Hi guys I have a few questions that I am totally lost on.

1) Write as a single loagrithm: 4lnx - 2(lnx^3 + 4lnx)

2) Solve: 2^x = 3^(x+3)

3) Solve: log(x^2 -1) = 2 + log (x+1)

Thanks in advance!!

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- Apr 15th 2008, 09:46 AM #1

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- Apr 15th 2008, 09:51 AM #2
what have you tried?

here are the rules you need:

$\displaystyle \log_a(x^n) = n \log_a x$

$\displaystyle \log_a(xy) = \log_ax + \log_a y$

$\displaystyle \log_a \left( \frac xy \right) = \log_a x - \log_a y$

2) Solve: 2^x = 3^(x+3)

3) Solve: log(x^2 -1) = 2 + log (x+1)

try them and see what you get

- Apr 15th 2008, 10:09 AM #3

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- Apr 15th 2008, 10:47 AM #4
$\displaystyle 4 \ln x - 2( \ln (x^3) + 4 \ln x) = \ln (x^4) - 2( \ln (x^3) + \ln (x^4))$ ..................since $\displaystyle n \log_a x = \log_a (x^n)$

$\displaystyle = \ln (x^4) - 2(\ln (x^7))$ ...................since $\displaystyle \log_a x + \log_a y = \log_a xy$

$\displaystyle = \ln (x^4) - \ln (x^{14})$ .....................since $\displaystyle n \log_a x = \log_a (x^n)$

$\displaystyle = \ln \left( \frac {x^4}{x^{14}}\right)$ .............................since $\displaystyle \log_a x - \log_a y = \log_a \frac xy$

$\displaystyle = \ln (x^{-10})$

2) Solve: 2^x = 3^(x+3)

$\displaystyle \Rightarrow \ln 2^x = \ln 3^{x + 3}$ ......................use the power rule for logs

$\displaystyle \Rightarrow x \ln 2 = (x + 3) \ln 3$ .................distribute

$\displaystyle \Rightarrow x \ln 2 = x \ln 3 + 3 \ln 3$ ...............get all x's to one side

$\displaystyle \Rightarrow x \ln 2 - x \ln 3 = \ln 27$ .................factor out the x

$\displaystyle \Rightarrow x (\ln 2 - \ln 3) = \ln 27$ .................solve for x

$\displaystyle \Rightarrow x = \frac {\ln 27}{\ln 2 - \ln 3}$ .....................simplify

$\displaystyle \Rightarrow x = \frac {\ln 27}{\ln \frac 23}$

3) Solve: log(x^2 -1) = 2 + log (x+1)

- Apr 15th 2008, 10:57 AM #5

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- Apr 15th 2008, 11:04 AM #6

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- Apr 15th 2008, 11:06 AM #7