Log 3 on base 2 * log 4 on base 3 * log 5 on base 4 ....... Log (n+1) on base n =10 find value of n
Answers
Answer:
here is your answer
Step-by-step explanation:
There is one other log "rule", but it's more of a formula than a rule.
You may have noticed that your calculator only has keys for figuring the values for the common (that is, the base-10) log and the natural (that is, the base-e) log. There are no keys for any other bases. Some students try to get around this by "evaluating" something like "log3(6)" with the following keystrokes:
[LOG] [ 3 ] [ ( ] [ 6 ] [ ) ]
Of course, they then get the wrong answer, because the above actually (usually) calculates the value of "log10(3) × 6". This is not what had been intended.
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Change of Base Formula on MathHelp.com
Change of Base Formula
In order to evaluate a non-standard-base log, you have to use the Change-of-Base formula:
Change-of-Base Formula:
\log_{\color{blue}{b}}(\color{red}{x}) = \dfrac{\log_d(\color{red}{x})}{\log_d(\color{blue}{b})}log
b
(x)=
log
d
(b)
log
d
(x)
What this rule says, in practical terms, is that you can evaluate a non-standard-base log by converting it to the fraction of the form "(standard-base log of the argument) divided by (same-standard-base log of the non-standard-base)". I keep this straight by looking at the position of things. In the original log, the argument is "above" the base (since the base is subscripted), so I leave things that way when I split them up:
argument "x" in top log, base "b" in bottom log
Here's a simple example of this formula's application:
Evaluate log3(6). Round your answer to three decimal places.
The argument is 6 and the base is 3. I'll plug them into the change-of-base formula, using the natural log as my new-base log:
\log_3(6) = \dfrac{\color{red}{\ln(6)}}{\color{blue}{\ln(3)}}log
3
(6)=
ln(3)
ln(6)
= \dfrac{\color{red}{1.79175946923...}}{\color{blue}{1.09861228867...}}=
1.09861228867...
1.79175946923...
= 1.63092975357...=1.63092975357...
Then the answer, rounded to three decimal places, is:
log3(6) = 1.631
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I would have gotten the same final answer if I had used the common log instead of the natural log, though the numerator and denominator of the intermediate fraction would have been different from what I displayed above:
\log_3(6) = \dfrac{\color{red}{\log(6)}}{\color{blue}{\log(3)}}log
3
(6)=
log(3)
log(6)
= \dfrac{\color{red}{0.778151250384...}}{\color{blue}{0.47712125472...}}=
0.47712125472...
0.778151250384...
= 1.63092975357...=1.63092975357...
As you can see, it doesn't matter which standard-base log you use, as long as you use the same base for both the numerator and the denominator.
While I showed the numerator and denominator values in the above calculations, it is actually best to do the calculations entirely within your calculator. You don't need to bother with writing out that intermediate step.
In fact, to minimize on round-off errors, it is best to try to do all the steps for the division and evaluation in your calculator, all in one go. In the above computation, rather than writing down the first eight or so decimal places in the values of ln(6) and ln(3) and then dividing, you would just do "ln(6) ÷ ln(3)" in your calculator.
Answer:
answer is 1023 because there is n+1 terms
