Algorithm change formula problem

[Illvoices]

SO how do solve for the following logarithm using the change formula?
log₃16

To solve i divide the 16 and 3 which gave me 5.333333

when the actual answer is 2.523719

 

[ksdhart2]

Okay, so I'm not sure I fully follow your post, but it looks to me like you're trying to use the change-of-base formula for logarithms to change this problem into a logarithm base-e (aka natural log), to make it easier to solve. Assuming that's the case, let's try reviewing exactly what the change-of-base formulahttp://www.mathwords.com/c/change_of_base_formula.htm says.

Change of base formula: $\displaystyle log_a(x)=\dfrac{log_b(x)}{log_b(a)}$

where a is the base of the original logarithm, x is the argument of the original logarithm, and b is the target base for the new logarithm. Typically this is the mathematical constant e, although the formula will work for any positive number b.

Based on this review, I believe you misunderstood the formula, believing it to be this: $\displaystyle log_a(x)=\dfrac{x}{a}$. But, as you can see that's not at all the case. I see you post a lot here, and that's perfectly fine, but I can't help but get the feeling that you're not really taking our advice to heart. I know I've personally advised you multiple times to really learn what the formula says, and make sure you really understand what it means before trying to use it. I can definitely see that you're trying very very hard to use all these formulas and rules, but when you don't even understand what they're doing they just become these magical tools that give you right answer if you use them correctly but you never actually learned the underlying concept, and so of course you're confused as all get out when the next concept tries to build on the knowledge you're supposed to have.

I think you might find this letterhttp://mathforum.org/library/drmath/view/70648.html written to "Ask Dr. Math" to be helpful. A student there asked why the change-of-base formula works and Dr. Math gave some insight as how you might derive it if it weren't already given to you by your teacher/textbook. Good luck.

 

[HallsofIvy]

SO how do solve for the following logarithm using the change formula?
log₃16

To solve i divide the 16 and 3 which gave me 5.333333

when the actual answer is 2.523719

Do you not know what the "change formula" is? There is no formula that says "$\displaystyle log_a(b)= \frac{b}{a}$! "Dividing 16 by 3" simply ignores the logarithm. If $\displaystyle x= log_3(16)$, then $\displaystyle 16= 3^x$. Taking "log", to any base, of both sides gives $\displaystyle log(16)= log(3^x)= x log(3)$. Thus, $\displaystyle log_3(16)= x= \frac{log(16)}{log(3)}$. Again that is for any base.

In particular, if "log" is the natural logarithm (base e), then log(16)= 2.7726 and log(3)= 1.0986 so that $\displaystyle log_3(16)= \frac{2.7726}{1.0986}= 2.5237$.

If "log" is the common logarithm (base 10), then log(16)= 1.2041 and log(3)= 0.4771 so $\displaystyle log_3(16)= \frac{1.2041}{0.4771}= 2.5237$ as before.

 

[Illvoices]

ive got the answer thanks to you guys. By applying the change algorithm formula i was able to divide log16 and log3 and get the correct answer. =)