• May 8th 2010, 03:47 AM
different92
Without the use of a calculator.

1. e^ln0.1
2. log(base)2 3. log(base)3 4. log(base)4 8
3. e^log(base)e^2^9
4. ln(y+4)=5x+lnC
5. log(base)5(x+6)+log(base)5 (x+1)=1
Determine the domain of the ff:
f(x)=log(base)2 (xsquare-3x+2).

When i say log(base)2, I mean the no that goes underneath the log.
Thanks a lot!(Crying)
• May 8th 2010, 04:50 AM
e^(i*pi)
Quote:

Originally Posted by different92
Without the use of a calculator.

1. e^ln0.1
2. log(base)2 3. log(base)3 4. log(base)4 8
3. e^log(base)e^2^9
4. ln(y+4)=5x+lnC
5. log(base)5(x+6)+log(base)5 (x+1)=1
Determine the domain of the ff:
f(x)=log(base)2 (xsquare-3x+2).

When i say log(base)2, I mean the no that goes underneath the log.
Thanks a lot!(Crying)

Some hints

Inverse functions: $\displaystyle e^{\ln(a)} = \ln(e^a) = a$

^ in general $\displaystyle \log_a(a^k) = a^{\log_a(k)} = k$

Change of base rule: $\displaystyle \log_b(a) = \frac{\log_c(a)}{\log_c(b)}$

Addition Rule: $\displaystyle \log_b(a) + \log_b(c) = \log_b(ac)$

Subtraction Rule: $\displaystyle \log_b(a) - \log_b(c) = \log_b\left(\frac{a}{c}\right)$

Power Law: $\displaystyle \log_a(b^k) = k\,\log_a(b)$
---------------------------

For example number 2 (I apologise if I have misread the question but the syntax is highly confusing)

From the change of base rule applied to each term (I have used base $\displaystyle e$ but it works with any positive base that is not 1): $\displaystyle \log_2(3) \cdot \log_3(4) \cdot \log_4(8) = \frac{\ln(3)}{\ln(2)} \cdot \frac{\ln(4)}{\ln(3)} \cdot \frac{\ln(8)}{\ln(4)}$

since $\displaystyle 8 = 2^3$ and $\displaystyle 4 = 2^2$ we can simplify using the power law

$\displaystyle \frac{\ln(3)}{\ln(2)} \cdot \frac{2\ln(2)}{\ln(3)} \cdot \frac{3\ln(2)}{2\ln(2)}$

Cancel out terms as you would with any ordinary fraction to give an answer of $\displaystyle 3$