someone answer please!!!
PLEASE PLEASE PLEASE
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
I know this is a lot of questions! PLEASE HELP! I'm desperate!! Thank You VERY VERY much in advance!!!!!!
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.
If you get stuck, please reply showing your steps and reasoning so far.
Thank you!
In order to change from base "a" to base "b" the following formula must be used:
for example:
<--- btw this means log 10 I'm still quite noob at the coding of it.
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.
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:
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...
Again, I leave the rest to the OP.
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