Daily Mechanics Homework
When presented with a problem involving a base other than "10" or "e", such as the following:
"2 = 3 ^ x",
what is the best strategy for arriving at the value of, x, assuming there is a "best strategy?
Apply this strategy and solve the following problems:
1. 3 ^ x = 8
2. 4 ^ y = 10
3. 2.1 ^ z = 100
4. 1,030 = 3.2 ^ s
5. 0.5 ^ t = 21
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log[10]x
ReplyDeletelog[2]x = ----------
log[10]2
This is the general equation for switching bases.
For these problems you can use the change of base formula, or do what i do since i don't like the change of base formula. You take log of both sides, and put the x out in front of the not ten base, then divide the log(whatever) away from the x and solve. Does that work? I used the log rule for exponents that says log3^x=xlog3
my answers were
1. x=1.893
2. y=1.661
3. z=6.644
4. s=5.964
5. t=-4.4
I am not 100% sure this is correct.
ReplyDeleteAll answers are rounded to two decimal places. Unfortunately I cannot use superscripts in a reply. Only in the original post.
ReplyDelete1. 3 ^ x = 8
Answer: X = 1.89
2. 4 ^ y = 10
Answer: Y = 1.66
3. 2.1 ^ z = 100
Answer: Z = 6.21
4. 1,030 = 3.2 ^ s
Answer: S = 5.96
5. 0.5 ^ t = 21
Answer: T = -4.39
we got the same essentially so i feel good about my work
ReplyDeleteExcellent work gentlemen and a good explanation of the method.
ReplyDeleteHow did you get 6.6 for the 3rd question? I know that Log(100) is 2 because 10^2 = 100 and log is base ten. So I did 2 / Log (2.1).
ReplyDeleteThe accepted death of time is before appearing mortis, which means the body hardens. Usually less than one hour will be appearing mortis. So if a person dead at 7:00 am,then the accepted death of time is before 7:59 am.
ReplyDeleteI also got the same answers:
ReplyDelete1. x=1.893
2. y=1.661
3. z=6.207
4. s=5.964
5. t=-4.392
I was pretty confused at first so I did some research and used the formula- (log base a)(x)=(log(x))/(log(a))-
this made things much clearer for me.
I got the same answers as everyone else, probably because Tashi helped me by teaching me that formula.
ReplyDeleteUsing the formula of changing bases:
ReplyDeletelog(a)(b)=log(c)(b)/log(c)(a)
(a,c>=0 and not equal to 1）
for example:
log(3)(4)=log(a)(4)/log(a)(3)
Because "a" can be anything,so everything can be changed to the form of lg or ln.
2 = 3 ^ x
x=log(3)(2)
x=lg(2)/lg(3)=0.631
---------------------------------------------
1)3 ^ x = 8
x=log(3)(8)=lg(8)/lg(3)=1.893
2)4 ^ y = 10
y=log(4)(10)=lg(10)/lg(4)=1.661
3)2.1 ^ z = 100
z=log(2.1)(100)=lg(100)/lg(2.1)=6.207
4)1,030 = 3.2 ^ s
s=log(3.2)(1030)=lg(1030)/lg(3.2)=5.964
5)0.5 ^ t = 21
t=log(0.5)(21)=lg(21)/lg(0.5)=-4.392
For Vivien, Jane and Tashi: Refer back to the original question and ask if your appriach is equivalent. So, are we simply using a property and, if so, how is this related to the change of base formula?
ReplyDeleteYes, I believe it is related. It is also equivalent to Julian's strategy just slightly simplified. I just skipped to the last step in the formula or approach.
ReplyDeleteThats what I started doing once I got the hang of it.
DeleteTo certificate the formula of changing bases:
ReplyDeleteIn log(a)(b)
assume that: a=n^x,b=n^y
Then,
log(a)(b)=log(n^x)(n^y)
According to these 2 formulas:
log(a）(M^n)=nloga(M)
and
log(a^n)M=1/n×log(a) M
So,
log(n^x)(n^y)=y/x
Because
a=n^x,b=n^y
so
x=log(n)(a),y=log(n)(b)
log(a)(b)=log(n^x)(n^y)=log(n)(b)/log(n)(a)