1. If log_a X=2, what is log_a (1/x) ?

2. Evaluate log_2 4 + log_4 8 + log_8 16 + log_16 32 + log_32 64 + log64 128

3. Simpilfy Expression :

x= (log_10 11)(log_11 12)....(log_98 99)(log_99 100)

4. log_2 3√ 16
(the 3 is a script, not 3√ 16)

5. (√ x) ^ log_x 2

This -----> _ is a subscript

3. Originally Posted by tecktonikk
1. If log_a X=2, what is log_a (1/x) ?

$\displaystyle \textcolor{red}{\log_a{x} = -\log_a\left(\frac{1}{x}\right)}$

2. Evaluate log_2 4 + log_4 8 + log_8 16 + log_16 32 + log_32 64 + log64 128

change of base formula, for example ...

$\displaystyle \textcolor{red}{\log_4(8) = \frac{\log_2(8)}{\log_2(4)}}$

3. Simpilfy Expression :

x= (log_10 11)(log_11 12)....(log_98 99)(log_99 100)

again, use the change of base formula ...

$\displaystyle \textcolor{red}{\log_{11}(12) = \frac{\log_{10}(12)}{\log_{10}(11)}}$

4. log_2 3√ 16
(the 3 is a script, not 3√ 16)

hint ... $\displaystyle \textcolor{red}{\sqrt[3]{16} = \sqrt[3]{2^4} = 2^{\frac{4}{3}}}$

5. (√ x) ^ log_x 2

two hints ... $\displaystyle \textcolor{red}{\sqrt{x} = x^{\frac{1}{2}}}$

$\displaystyle \textcolor{red}{x^{\log_x{2}} = 2}$
...

4. 1) How is "X" related to "x"? (The exercise, as currently posted, doesn't specify how these two different variables relate.)

2) Apply the definition of logarithms. What power must you put on "2" to get "4"? This power is the numerical value of log_2(4). And so forth.

3) Convert each log to base-10, using the change-of-base formula, and cancel off the common factors in the result. The answer is surprising simple.

4) Express the cube root of sixteen as a power of 2, and then apply the definition.

5) First, convert the square root to a fractional power. Apply an exponent rule to reverse the order of the powers. Apply a form of the definition of logs to simplify the expression inside the parentheses. Then apply the square root to the result.

Thank you!

5. I apologize... I still don't get it.

Even if I change the base formula...

how do I add them together??

6. Originally Posted by tecktonikk
I apologize... I still don't get it.

Even if I change the base formula...

how do I add them together??
Are you referring to exercise (3)? What would you be adding?

7. I'm referring to question # 2

8. Originally Posted by tecktonikk
I'm referring to question # 2

using the change of base formula, get each term in the form of a fraction without a logarithm, then add.

9. In order to change from base "a" to base "b" the following formula must be used:

$\displaystyle log_a x = \frac {log_b x}{log_b a}$

for example:

$\displaystyle log_6 53$

$\displaystyle = log_6 53 = \frac {log_10 53}{log_10 6}$ <--- btw this means log 10 I'm still quite noob at the coding of it.

$\displaystyle = \frac {1.7242759}{0.7781512}$

$\displaystyle = 2.22$

hope this clears things up because this formula is quite important in my opinion...

so since your equation is telling you to add different bases, it is easier to convert all those bases into base 10 (normal numbers) then just add using simple arithmetic.

10. Originally Posted by jgv115
$\displaystyle = log_6 53 = \frac {log_10 53}{log_10 6}$ <--- btw this means log 10 I'm still quite noob at the coding of it.
The underscore "_" subscripts things in LaTeX; the carat "^" superscripts things. But only the one thing immediately following the symbol. To put the "10" into the subscript, use curly braces "{}".

[php]$\displaystyle \log_{10}(53)$[/php]

...displays as:

. . . . .$\displaystyle \log_{10}(53)$

Have fun!

11. tecktonikk, begging and write in such a huge font like that is kinda rude, and you're less likely to get help that way.

Originally Posted by jgv115
so since your equation is telling you to add different bases, it is easier to convert all those bases into base 10 (normal numbers) then just add using simple arithmetic.
I disagree... I think all of the logs should be in base 2.

2) $\displaystyle \log_{2} 4 + \log_{4} 8 + \log_{8} 16 + \log_{16} 32 + \log_{32} 64 + \log_{64} 128$

Use the change-of-base formula like this:
$\displaystyle = \log_{2} 4 + \frac{\log_{2} 8}{\log_{2} 4} + \frac{\log_{2} 16}{\log_{2} 8}+ ...$

I'll leave the rest to the OP.

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12. If you make it base 2 then you still have to convert to base 10 in the end?

13. Originally Posted by tecktonikk
2. Evaluate log_2 4 + log_4 8 + log_8 16 + log_16 32 + log_32 64 + log64 128
recommend base 2 instead of base 10 for the following reason ...

$\displaystyle \log_2{4} + \frac{\log_2{8}}{\log_2{4}} + \frac{\log_2{16}}{\log_2{8}} + \frac{\log_2{32}}{\log_2{16}} + \frac{\log_2{64}}{\log_2{32}} + \frac{\log_2{128}}{\log_2{64}}$

$\displaystyle 2 + \frac{3}{2} + \frac{4}{3} + \frac{5}{4} + \frac{6}{5} + \frac{7}{6}$

14. alright. i'm not very good. I guess I learnt something

15. Originally Posted by jgv115
alright. i'm not very good. I guess I learnt something
Don't fret. It's all about noticing the little things, like in #2 all of the numbers involved are powers of 2, so keeping the logarithms in base 2 is the best.

Now in #3, it does make sense to keep it in base 10:
Originally Posted by tecktonikk
3. Simpilfy Expression :

x= (log_10 11)(log_11 12)....(log_98 99)(log_99 100)
Disclaimer: where I live, a log without a base is assumed to be a log base 10.

We use the change-of-base formula again:
x...
$\displaystyle = (\log_{10} 11)(\log_{11} 12)(\log_{12} 13)....(\log_{98} 99)(\log_{99} 100)$

$\displaystyle = \log 11 \left(\frac{\log 12}{\log 11}\right)\left(\frac{\log 13}{\log 12}\right)...$

Again, I leave the rest to the OP.

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