The first sequence looked like this:
log2^123 = 3/1
log2^223 = 3/2
log2^323 = 3/3
log2^423 = 3/4
log2^523 = 3/5
:
:
log2^n23 = 3/n
The nth term of this sequence can be expressed as 3/n.
I completed the same steps to discover the general expressions of each of the following sequences.
log381, log981, log2781, log8181, ... can be written log3^134, log3^234, log3^334, log3^434, …
log3^134 = 4/1
log3^234 = 4/2
log3^334 = 4/3
log3^434 = 4/4
:
:
log3^n34 = 4/n
The nth term of the second sequence can be expressed as 4/n.
log525, log2525, log12525, log62525, … can be written log5^125, log5^225, log5^325, log5^425, …
log5^152 = 2/1
log5^252 = 2/2
log5^352 = 2/3
log5^452 = 2/4
:
:
log5^n52 = 2/n
The nth term of the third sequence can expressed as 2/n.
The same rules apply for the fourth and final sequence.
logm^1mk, logm^2mk, logm^3mk, logm^4mk, …
logm^1mk = k/1
logm^2 mk = k/2
logm^3 mk = k/3
logm^4 mk = k/4
:
:
logm^n mk = k/n
The nth term of the fourth sequence can expressed as k/n.
Now, calculate the following, giving your answers in the form p/q, where p and q are both integers.
log464, log864, log3264 3/1, 2/1, 6/5
log749, log4949, log34349 2/1, 2/2, 2/3
log1/5125, log1/125125, log1/625125 3/-1, 3/-3, 3/-4
log8512, log2512, log16512 3/1, 9/1, 9/4
Describe how to obtain the third answer in each row from the first two answers.
First, I took the lowest prime base possible of each.
log2^264, log2^364, log2^564
log7^149, log7^249, log7^349
log(1/5)^1125, log(1/5)^3125, log(1/5)^4125
log2^3512, log2^1512, log2^4512
In doing this, I noticed a pattern : the exponent of the base of the answer in the third column is simply the sum of the exponents of the bases in the first two columns when each base is in its lowest prime form. (expbase1+ expbase2 = expbase3 )
log2^264, log2^364, log2^564 (2+3=5)
log7^149, log7^249, log7^349 (1+2=3)
log(1/5)^1125, log(1/5)^3125, log(1/5)^4125 (1+3=4)
log2^3512, log2^1512, log2^4512 (3+1=4)
Create two more examples that fit the pattern above.
log4256, log16256, log64256
(4) (2) (4/3)
log2^2256 log2^4256 log2^6256 (2+4=6)
log9729, log3729, log27729
(3) (6) (2)
log3^2729 log3^1729 log3^3729 (2+1=3)
Let logax = c and logbx = d. Find the general statement that expresses logabx, in terms of c and d.
First,
logax=c can otherwise be written x= ac , and logbx=d can otherwise be written x= bd
Next, a and b can be isolated.
x1/c=a x1/d=b
They can then be multiplied, as the goal is to find logabx.
x1/c+1/d=ab
The variable x can now be isolated.
x=(ab)1/(1/c+1/d)
The irrational fraction in the exponent can be eradicated my multiplying by 1, which can be written as cd/cd.
x=(ab)cd/(c+d)
This expression can be rewritten now, as
logabx= cd/(c+d)
Thus, the general statement for logabx in terms of c and d is logabx= cd/(c+d).
Test the validity of your general statement using other values of a, b, and x.
All right, suppose a= 3, b=9, and x=81. The expression still holds true.
logax=c logbx=d
log381=4 log981=2
c=4 d=2
logabx= cd/(c+d)
log2781= 8/6= 4/3
or
274/3=81
Or suppose a=8, b=4, and x= 64. The expression is consistent.
logax=c logbx=d
log864=4 log464=2
c=4 d=3
logabx= cd/(c+d)
log3264= 6/5
or
326/5=64
Discuss the scope and/or limitations of a, b and x.
The variables a and b must be integers greater than zero and not equal to one. If a or b were equal to one, then c and d would be rendered undefined.
(For example, if a=1 and x=3, the given statement logax=c would be undefined for c
because there is no exponent that can be raised to make one equal three. One to any power always equals one.)
This is the same for b. If it were equal to one, d would be undefined.
They cannot be negative, as this does not work with the equations.
The variable c cannot equal negative d, as this would yield a zero in the denominator of the equation, which would make it undefined.
Explain how you arrived at your general statement.
I explained it step by step above, but generally I just used my knowledge of logarithm and exponent rules to mold and simplify the information I was given. For example, I knew that logax=c was the same as writing x= ac, a point I explained in the brief introduction, which helped me not only with this portion but the portfolio in its entirety.
