1)given that log(y)/log(4)=a and log(2y)/log(8)=b, show that 2a=3b-1
2)the strength (S) of an earthquake is represented by ritcher scale (R) and the relationship is as follows:
R=log(S)/log(10).show that when the ritcher scale shows an increase in value by 1,the strength of the earthquake will have increased by 10 times.
Note that log_2(n) = 2*log_4(n) = 3*log_8(n)
a = log_4(y) = (1/2) log_2(y)
b = log_8(2y) = log_8(2) + log_8(y) = 1/3 + (1/3) log_2(y)
b = 1/3 + (2/3) a
3b = 1+2a
If logn = logm + 1,
logn = logm + log10 = log(10m)
so, n = 10m
1) Oh boy, let's put on our math hats! Alright, here we go. Starting with the first equation: log(y)/log(4)=a. We can rearrange this equation by multiplying both sides by log(4): log(y) = a * log(4).
Next, for the second equation: log(2y)/log(8)=b. To simplify things, let's use the change of base formula to convert the denominator to log(2). So we have: log(2y)/(log(8)/log(2))=b. That means we can rewrite this equation as: log(2y)*log(2)/log(8)=b.
Now, let's simplify the second equation further by writing log(8) as 3 * log(2), since 8 is equal to 2 raised to the power of 3. So, we have: log(2y)*log(2)/(3*log(2))=b. Cancelling out log(2) on the numerator and denominator, we get: log(2y)/3=b.
Now, let's go back to the first equation: log(y) = a * log(4). If we apply the change of base formula here as well, we can rewrite it as: log(y)*log(2)/log(4)=a. But what is log(4)? It's 2*log(2)! So, we can simplify further: log(y)*log(2)/(2*log(2))=a. Cancellling out log(2) on the numerator and denominator, we get: log(y)/2=a.
Okay, now we have log(y)/2=a and log(2y)/3=b. Let's multiply the first equation by 2 and the second equation by 3 to make the coefficients the same: 2 * log(y)/2 = 2a, and 3 * log(2y)/3 = 3b.
Simplifying those expressions, we get: log(y) = 2a and log(2y) = 3b. Notice anything? We can rewrite log(2y) as log(2) + log(y) using the log properties. So, we have: 2a = log(y) and 3b = log(2) + log(y).
Now here comes the grand finale! If we subtract log(2) from both sides of the second equation, we have 3b - log(2) = log(y). And guess what? That's the same as 2a! So we can conclude that 2a = 3b - log(2).
2) Ah, earthquakes! Well, the Richter scale is quite remarkable. So, we have the relationship R = log(S)/log(10). Let's see what happens when R increases by 1, shall we?
If we start with the original equation and add 1 to both sides, we get: R + 1 = log(S)/log(10) + 1. And because log(10) is equal to 1, we can simplify this to: R + 1 = log(S) + 1.
Now, let's exponentiate both sides with a base of 10 to get rid of the logarithm. We end up with: 10^(R + 1) = 10^(log(S) + 1). Exponentiating a logarithm with its base eliminates the logarithm, so it becomes: 10^(R + 1) = S * 10^1, which simplifies to 10^(R + 1) = 10S.
Okay, almost there! We can rewrite 10^(R + 1) as 10 * 10^R, since that's how exponential notation works. So, we have: 10 * 10^R = 10S.
And remember, 10^R is just S! So, we get: 10S = 10S. You see? When R increases by 1, the strength of the earthquake (S) remains the same, but it's represented on a different scale. So in terms of the Richter scale, an increase of 1 means the earthquake strength has increased by 10 times! Mind-blowing, isn't it?
1) Start with the given equations:
log(y)/log(4) = a ----(1)
log(2y)/log(8) = b ----(2)
To simplify the equations, we can rewrite the logarithms using the change of base formula:
log(y) / log(4) = a
log(y) / log(2^2) = a
log(y) / (2 * log(2)) = a
(log(y) / log(2)) / 2 = a
(1/2) * log(y) / log(2) = a
Similarly,
log(2y) / log(8) = b
log(2y) / log(2^3) = b
log(2y) / (3 * log(2)) = b
(1/3) * log(2y) / log(2) = b
Now, let's solve for log(y) from equation (1):
(1/2) * log(y) / log(2) = a
log(y) / log(2) = 2a
log(y) = 2a * log(2)
log(y) = log(2^(2a))
Since the logarithms are equal, we can equate their arguments:
y = 2^(2a) ----(3)
Similarly, solving for log(2y) from equation (2):
(1/3) * log(2y) / log(2) = b
log(2y) / log(2) = 3b
log(2y) = 3b * log(2)
log(2y) = log(2^(3b))
Again, equating the logarithms:
2y = 2^(3b) ----(4)
Now, let's solve for 2a using equation (3):
y = 2^(2a)
y = 2^2a
Taking the logarithm base 2 of both sides:
log(y)/log(2) = 2a * log(2) / log(2)
log(y) = 2a * log(2)
2a = log(y) / log(2) ----(5)
Similarly, solving for 3b using equation (4):
2y = 2^(3b)
Taking the logarithm base 2 of both sides:
log(2y) / log(2) = 3b * log(2) / log(2)
log(2y) = 3b * log(2)
3b = log(2y) / log(2) ----(6)
Substituting equations (5) and (6) into the expression 2a = 3b - 1:
2a = 3b - 1
log(y) / log(2) = (log(2y) / log(2)) - 1
Multiplying through by log(2) to clear the denominators:
log(y) = log(2y) - log(2)
log(y) = log(2y/2)
log(y) = log(y)
The equation holds true, so we have verified that 2a = 3b - 1.
Therefore, we have shown that 2a = 3b - 1.
2) Given the relationship between the Richter scale (R) and earthquake strength (S): R = log(S) / log(10).
Let's compare two scenarios where the Richter scale increases by 1 and the earthquake strength increases by x times.
Scenario 1: Increase in Richter scale by 1
R1 = log(S1) / log(10)
Scenario 2: Increase in earthquake strength by x times
R2 = log(S2) / log(10)
According to the problem, the increase in Richter scale is always by 1, so we can express R2 in terms of R1 as:
R2 = R1 + 1
Now, let's solve for the corresponding earthquake strengths S1 and S2 using the given equations and relationship:
R1 = log(S1) / log(10) ----(1)
R2 = log(S2) / log(10) ----(2)
Substituting R2 = R1 + 1 in equation (2):
R1 + 1 = log(S2) / log(10)
Multiplying through by log(10) to clear the denominators:
log(10) * (R1 + 1) = log(S2)
Using the logarithmic property log(a * b) = log(a) + log(b):
log(10^(R1 + 1)) = log(S2)
According to the definition of a logarithm, log(a^b) = b * log(a):
(R1 + 1) * log(10) = log(S2)
Since log(10) = 1 (log of the base itself):
(R1 + 1) = log(S2)
Now, let's solve for the earthquake strength S1 using equation (1):
R1 = log(S1) / log(10)
R1 * log(10) = log(S1)
Since log(10) = 1, we have
R1 * 1 = log(S1)
R1 = log(S1)
Comparing the expressions for R1 and R1 + 1 from the two scenarios:
R1 = log(S1)
R1 + 1 = log(S2)
We can see that the earthquake strength S2 is obtained by raising 10 to the power of (R1 + 1), while S1 is obtained by raising 10 to the power of R1:
S2 = 10^(R1 + 1)
S1 = 10^R1
Dividing S2 by S1:
(S2 / S1) = (10^(R1 + 1)) / (10^R1)
(S2 / S1) = 10^1 (Using the property, a^m / a^n = a^(m-n))
Since 10^1 = 10, we can conclude that when the Richter scale shows an increase in value by 1, the strength of the earthquake will have increased by 10 times.
Therefore, we have shown that when the Richter scale shows an increase in value by 1, the strength of the earthquake will have increased by 10 times.
1) To show that 2a = 3b - 1 using the given logarithmic equations, let's simplify each equation separately and then substitute the results into the final equation.
Starting with the first equation:
log(y)/log(4) = a
We can apply the "change of base formula" to convert log(4) to another base. Let's convert it to log(2):
log(y)/log(2) / log(4)/log(2) = a
Using the properties of logarithms, log(a)/log(b) is equal to log_b(a), so we can rewrite it as:
log(y)/log(2) = a
Similarly, for the second equation:
log(2y)/log(8) = b
Using the change of base formula again, but this time converting log(8) to log(2):
log(2y)/log(2) / log(8)/log(2) = b
Again, applying log(a)/log(b) = log_b(a), we get:
log(2y)/log(2) = b
Now, let's simplify further:
log(y)/log(2) = a --(1)
log(2y)/log(2) = b --(2)
From equation (1), we can rewrite it as:
log(y) = a * log(2)
And from equation (2), we can rewrite it as:
log(2y) = b * log(2)
Now, let's manipulate equation (2):
log(2y) = b * log(2)
log(2y) = log(2^b) (using the power rule of logarithms)
2y = 2^b (using the definition of a logarithm)
Now, let's manipulate equation (1):
log(y) = a * log(2)
log(y) = log(2^a) (using the power rule of logarithms)
y = 2^a (using the definition of a logarithm)
Now, we can substitute these values back into the final equation to prove the given equality:
2^a = 2y (from equation (1))
2^a = 2 * 2^b (substituting y from equation (2) into the equation)
2^a = 2^(b+1) (using the property of exponents)
a = b + 1 (taking the logarithm of both sides)
Now, let's manipulate this equation to match the desired form of 2a = 3b - 1:
2a = 2b + 2 (multiply both sides by 2)
2a - 2 = 2b (subtract 2 from both sides)
2(a - 1) = 2b (factor out 2 on the left side)
a - 1 = b (divide both sides by 2)
Substituting this back into the equation 2a = 3b - 1:
2(a - 1) = 3b - 1
2a - 2 = 3b - 1
2a = 3b - 1
Therefore, we have shown that 2a = 3b - 1 using the given logarithmic equations.
2) To show that when the Richter scale shows an increase in value by 1, the strength of the earthquake will have increased by 10 times using the given relationship R = log(S)/log(10), we need to compare the strengths of two earthquakes that differ by 1 on the Richter scale.
Let's assume there are two earthquakes with Richter scale values R1 and R2, where R1 = R2 + 1.
Using the given relationship, we have:
R1 = log(S1)/log(10)
R2 = log(S2)/log(10)
Now, let's compare the strengths of the earthquakes:
R1 - R2 = log(S1)/log(10) - log(S2)/log(10)
R1 - R2 = [log(S1) - log(S2)]/log(10) (using the property log(a) - log(b) = log(a/b))
R1 - R2 = log(S1/S2)/log(10) (using the property log(a) - log(b) = log(a/b))
Since R1 = R2 + 1, we can substitute it:
1 = log(S1/S2)/log(10)
log(10) = log(S1/S2) (multiplying both sides by log(10))
Now, let's convert it to exponential form using the definition of logarithms:
10 = S1/S2
From this equation, we can conclude that when the Richter scale shows an increase in value by 1, the strength of the earthquake will have increased by 10 times.
Therefore, we have shown that when the Richter scale shows an increase in value by 1, the strength of the earthquake will have increased by 10 times using the given relationship R = log(S)/log(10).