Lorena T.
asked • 02/01/23Given m=log a, n= log b, and k= log ab
a. express a, b, and ab as powers of 10
b. with knowledge of exponents what is the relationship among m, n, and k
c. conclude that log (ab) = log a + log b
Justify rules:
a. log (a^r) = r log a
b. log (a/b) = log a - log b
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2 Answers By Expert Tutors
Sean R. answered • 02/02/23
Tutor
5
(15)
Experienced MCAT, Physio, Bio., Chem, Stats, & Math Tutor from UCLA
(a) 10^m = 10log10(a). The 10log10 cancels out b/c the definition of logarithmic functions is to the power 10; logarithmic functions are the inverse functions of exponentiation, or exponents, such that:
logb(bx) = x.
Given this relationship, we can rearrange variables with values given in the presented question stem whereby a, b, and ab may be expressed as powers of 10. Lets say we set the base of the logarithm, "b" in the image aforementioned. Then log10(10x) = x, because the " log10(10" term on the left side of the equation cancels out, leaving only the x remaining.
Given that m = log(a), we can presume the base of the logarithmic function to be 10, such that exponentiating both sides of the equation with 10^(***), we can cancel out the logarithmic function (since logs are the inverse functions of exponentiation). The following is what comes from exponentiating both sides by the base factor of 10:
10m = 10log10(a) . The 10log10 of course cancels out, leaving us with 10m = a.
n = log10(b) same thing leading to 10n = b.
k = log10(ab) : 10k = ab.
(b) Knowledge of exponents, the fact that log functions are the inverse of exponentiation functions, and that we can combine terms following simplifications, produces the following proof and derivation:
ab = 10k = (10m)(10n). Because the bases, 10, are the same in each factor undergoing multiplication on the right side of the equation, the exponents may be summed which equates to the terms being multiplied. This leaves us with:
10k = 10m + n. Given that the bases of 10 are equivalent on both sides of the equation, we can equate the exponents to each other leaving us with:
k = m + n.
(c) Coming full circle, if we take the log of both side of the 10k = ab function, we get the following:
log1010k = log10(ab). Simplifying:
k = log10(ab). As we derived up above, k = m + n.
m = log10(a)
n = log10(b).
So, k = log10(ab) = log10(a) + log10(b). This concludes the proof with the final (c) described in the question stem.
Sean R.
tutor
I was not able to fit all the proofs in only one comment, so here are the remaining justifications for (a) and (b):
Justify rules:
a. log (a^r) = r log a
b. log (a/b) = log a - log b
Justifying (a), log10(ar) = r log10(a):
Let us take m = log10(a).
So, log10(ar) = (r)(m).
Exponentiate both sides of the equation to the same value as in the base of the log10.
This leave us with:
10log10(ar) = 10(r)(m).
The 10^log cancels, leaving us with:
ar = 10(r)(m).
Now if I take the log of both sides, justifying log(ar), I am left with:
log10(ar) = log1010(r)(m).
Canceling the log10 10 on the right side gives:
log10(ar) = (r)(m).
Plugging back in what we set for m in the beginning of the proof gives:
log10(ar) = rlog10(a).
Justifying (b),
log(a/b) = log(a) - log(b):
let us take m = log10(a) and n = log10(b).
now if we exponentiate both sides to the base factor of the log's base functions of 10, we get:
10m = 10log10(a).
the 10log10 on the right side of the equation cancels, leaving us with:
10m = a
We would similarly get:
10n = b.
Now if you take a divided by b (for the a/b argument of the logarithmic function we are justifying), then we get:
a/b = 10m / 10n.
Since both bases are the same, 10, we can simplify this as:
a/b = 10m-n.
Now take the log of both sides:
log10(a/b) = log1010m-n.
This leaves us with:
log10(a/b) = m - n, where
m = log10(a) and n = log10(b).
Justifying: log10(a/b) = log10(a) - log10(b).
Please let me know if you have any remaining questions!
Report
02/02/23
Sean R.
tutor
I was not able to fit all the proofs in only one comment, so here are the remaining justifications for (a) and (b):
Justify rules:
a. log (a^r) = r log a
b. log (a/b) = log a - log b
Justifying (a), log10(ar) = r log10(a):
Let us take m = log10(a).
So, log10(ar) = (r)(m).
Exponentiate both sides of the equation to the same value as in the base of the log10.
This leave us with:
10log10(ar) = 10(r)(m).
The 10^log cancels, leaving us with:
ar = 10(r)(m).
Now if I take the log of both sides, justifying log(ar), I am left with:
log10(ar) = log1010(r)(m).
Canceling the log10 10 on the right side gives:
log10(ar) = (r)(m).
Plugging back in what we set for m in the beginning of the proof gives:
log10(ar) = rlog10(a).
Justifying (b),
log(a/b) = log(a) - log(b):
let us take m = log10(a) and n = log10(b).
now if we exponentiate both sides to the base factor of the log's base functions of 10, we get:
10m = 10log10(a).
the 10log10 on the right side of the equation cancels, leaving us with:
10m = a
We would similarly get:
10n = b.
Now if you take a divided by b (for the a/b argument of the logarithmic function we are justifying), then we get:
a/b = 10m / 10n.
Since both bases are the same, 10, we can simplify this as:
a/b = 10m-n.
Now take the log of both sides:
log10(a/b) = log1010m-n.
This leaves us with:
log10(a/b) = m - n, where
m = log10(a) and n = log10(b).
Justifying: log10(a/b) = log10(a) - log10(b).
Please let me know if you have any remaining questions!
Report
02/02/23
Sophia T. answered • 02/02/23
Tutor
New to Wyzant
Specialized Math & Physics Tutor, AP Expertise
a.
Our definition of log tells us that if m = log(a), then 10m = a. So, a = 10m, b = 10n, and ab = 10k.
b.
With exponent rules, ab = (10m)(10n) = 10m+n = 10k. For this equation to be true, m+n=k.
c.
If m+n=k, we can use our original log definitions to say that log(a) + log(b) = log(ab).
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Mark M.
Is there a question or just getting the assignment done?02/01/23