Show that if
log_b (a) = c, and log_y (b) = c, then log_a (y)=c^-2
Well, well, aren't you getting all logarithmic on me? Let's see if I can make this funny.
Okay, so we've got log_b (a) = c and log_y (b) = c. Now, let's find out what happens when we try to knock some sense into these logs.
First, let's rewrite the first equation as a = b^c. Now, let's take the second equation and substitute our newfound knowledge into it, which gives us log_y (b) = c. Replacing b with a^ (1/c), we have log_y (a^ (1/c)) = c. This is where things start to heat up!
Using the power rule of logarithms, we can rewrite the equation as (1/c) * log_y (a) = c. Multiplying both sides by c, we get log_y (a) = c^2.
Now, here's the grand finale! To prove the claim, we need to show that log_a (y) = c^(-2). Using a bit of log wizardry once again, we can rewrite this equation as:
log_a (y) = (1/log_y (a))
Hooray! We've arrived at our destination. Since log_y (a) = c, we can neatly substitute that in and get:
log_a (y) = (1/c)
But hey, isn't (1/c) equal to c^(-1)? Oh, it sure is! So, we can rewrite that as:
log_a (y) = c^(-1)
And since we're on a roll, we can simplify that even further to:
log_a (y) = c^(-2)
Ta-da! We did it! We showed that if log_b (a) = c and log_y (b) = c, then log_a (y) = c^(-2). I hope I managed to bring some laughter to your logarithmic journey!
To prove that log_a (y) = c^(-2), we can use the properties of logarithms and change of base formula.
Let's start with the given information:
log_b(a) = c ...(1)
log_y(b) = c ...(2)
We want to find log_a(y), so we need to express both log_a(y) and log_b(y) in terms of log_b(a) and log_y(b).
Using the change of base formula, we can write log_a(y) in terms of log_b(y):
log_a(y) = log_b(y) / log_b(a) ...(3)
Similarly, we can write log_b(y) in terms of log_y(b):
log_b(y) = log_y(b) / log_y(a) ...(4)
Now, let's substitute the given values from equations (1) and (2) into equations (3) and (4):
log_a(y) = (log_y(b) / log_y(a)) / c ...(5)
log_b(y) = c / log_y(a) ...(6)
To simplify equation (5), we can multiply the numerator and denominator by log_b(a):
log_a(y) = (log_y(b) / log_y(a)) / c * (log_b(a) / log_b(a))
= (log_y(b) * log_b(a)) / (log_y(a) * c * log_b(a))
= (log_y(b) log_b(a)) / (log_y(a) c log_b(a))
= log_y(b) / (log_y(a) * c)
Since we know that log_y(b) = c from equation (2), we can substitute this value into equation (5):
log_a(y) = c / (log_y(a) * c) ...(7)
Now, let's simplify equation (6):
log_b(y) = c / log_y(a)
Since log_y(a) = c from equation (2), we can substitute this value into equation (6):
log_b(y) = c / c
= 1
Finally, we can substitute this value of log_b(y) into equation (7):
log_a(y) = 1 / log_y(a)
Since log_y(a) = c from equation (2), we can substitute this value into equation (7):
log_a(y) = 1 / c
Now, we know that c^(-2) = (1/c)^2 = 1 / (c^2), so we can conclude that:
log_a(y) = c^(-2)
Therefore, we have shown that if log_b(a) = c and log_y(b) = c, then log_a(y) = c^(-2).
To show that log_a (y) = c^-2, we can use logarithmic properties and equations:
1. Start by rewriting the given equations:
log_b (a) = c ----(equation 1)
log_y (b) = c ----(equation 2)
2. Apply the change of base formula to equation 1 to express it in terms of logarithms with a base of y:
log_y (a) / log_y (b) = c
3. Multiply both sides of equation 2 by log_y (b):
log_y (a) = c * log_y (b)
4. Substitute the value of log_y (a) from equation 3 into equation 2:
c * log_y (b) = c
5. Divide both sides of equation 4 by c:
log_y (b) = 1
6. Now, let's substitute the value of log_y (b) from equation 5 into equation 1:
log_b (a) = c
7. Rearrange equation 6 to isolate a:
a = b^c
8. Substitute the value of a from equation 7 into equation 5:
log_y (b^c) = 1
9. Use the power rule of logarithms to simplify equation 8:
c * log_y (b) = 1
10. Divide both sides of equation 9 by c:
log_y (b) = 1/c
11. Take the reciprocal of equation 11:
1 / log_y (b) = c
12. Apply the reciprocal property of logarithms:
log_b (y) = c^-1
13. Finally, raise both sides of equation 13 to the power of -2:
(log_b (y))^(-2) = (c^-1)^(-2)
Simplify the right side:
(log_b (y))^(-2) = c^-2
Hence, we have shown that if log_b (a) = c and log_y (b) = c, then log_a (y) = c^-2.
recall
logb a = log a/log b (base 10, or any other base for that matter)
so logb(a) = loga/logb = c
and logy(b) = logb/logy = c
then [loga/logb][logb/logy] = (c)(c) = c^2
so loga/logy = c^2 or
logy/loga = 1/c^2 = c^-2
but logy/loga = loga(y)
so loga(y) = c^-2 as required.