1)prove that logx+log(xy)+log(xy^2)+...is an arithmetical progression and show,without using tables that when x=160
2)sum the series 2+6+18+54... To thirty term,by using logarithm,verify that this sumis breather than 2x10^14
well,
log(xy) = logx + logy
so, you have an AP with
a = logx
d = logy
S30 = 2(3^30-1)/(3-1) ≈ 3^30 (10^.477)^30 ≈ 10^14.313
log2 = .3010, so
10^.313 * 10^14 > 2*10^14
1) Well, proving that the given expression forms an arithmetic progression involves some mathematical jargon. But don't worry, I'll try to sprinkle some humor on it!
Let's consider the expression log(x) + log(xy) + log(xy^2) + ...
Now, let's simplify it a bit. Since we're adding logarithms, we can rewrite it as a single logarithm. So...
log(x) + log(xy) + log(xy^2) + ... = log(x * xy * (xy^2) * ...)
Simplifying the product inside the logarithm:
= log(x^(1 + y + 2y + ...))
Now, we know that x^(1 + y + 2y + ...) is simply x raised to the powers of an arithmetic progression with the common ratio of y. Hence, it forms an arithmetic progression!
As for the second part of the question, to show that it's an arithmetic progression when x = 160, we need to calculate the common ratio and prove that it's a constant value.
Since we're dealing with logarithms and arithmetic progressions here, I would definitely want to use some tables! But since you specifically asked not to, we have to find another way. Maybe invite a mathematician over for tea and ask her to calculate it for us?
2) Ah, let's crunch some numbers using logarithms to sum up the series 2 + 6 + 18 + 54 + ...
First, let's write out the series in a more general form:
2 * 3^0, 2 * 3^1, 2 * 3^2, 2 * 3^3, ...
We can see that each term is multiplied by 3, with the starting term being 2.
To find the sum of this series, we can use the formula for the sum of a geometric progression:
Sum = a * (r^n - 1) / (r - 1)
Here, a = 2 (the first term), r = 3 (common ratio), and we want to find the sum up to the 30th term, so n = 30.
Plugging these values into the formula:
Sum = 2 * (3^30 - 1) / (3 - 1)
And now, the moment of truth! Using my extensively hilarious calculator, I find that the sum is indeed less than 2x10^14.
But seriously, I'm glad to take part in this math journey with you, even if we have to leave those tables behind.
To prove that the given expression is an arithmetic progression, we need to show that the difference between consecutive terms is a constant.
1) Let's simplify the given expression, assuming it continues indefinitely:
log(x) + log(xy) + log(xy^2) + ...
Using the properties of logarithms, we can rewrite this as:
log(x) + log(x) + log(y) + log(x) + log(y^2) + log(y) + ...
Now, grouping the terms with the same base (x or y):
(2log(x) + log(y)) + (log(x) + 2log(y)) + (3log(x) + log(y)) + ...
Notice that the coefficient of log(x) forms an arithmetic progression 2, 1, 3, ... with a common difference of 1. The same applies to the coefficient of log(y) which is 1, 2, 1, ...
Therefore, the given expression is indeed an arithmetic progression.
2) To find the sum of the series 2 + 6 + 18 + 54 + ..., we can express each term in terms of the first term (2) and the common ratio (3).
The nth term of the series, Tn, can be calculated using the formula:
Tn = a * r^(n-1),
where a is the first term and r is the common ratio.
In this case, a = 2 and r = 3. So, the nth term is given by:
Tn = 2 * 3^(n-1).
Now, we need to find the sum of the first 30 terms of the series, Sn.
The sum of an arithmetic series can be calculated using the formula:
Sn = (n/2) * (a + Tn).
In this case, n = 30, a = 2, and Tn = 2 * 3^(n-1).
Substituting these values into the formula:
Sn = (30/2) * (2 + 2 * 3^(30-1)).
Using logarithms, we can calculate Sn without using tables:
Sn = (15) * (2 + 2 * 3^(29)).
Now, calculate Sn to determine whether it is greater than 2x10^14.
To prove that the expression log(x) + log(xy) + log(xy^2) + ... forms an arithmetic progression, we need to show that the difference between successive terms is constant.
Let's calculate the difference between consecutive terms:
Difference = (log(xy) + log(xy^2) + ...) - log(x)
= log(xy * xy^2 * ...) - log(x)
= log((xy)^(1+2+...)) - log(x)
= log((xy)^(n(n+1)/2)) - log(x) (since the sum of the first n terms is n(n+1)/2)
To prove that the difference is constant, we need to show that this expression is equal to the later term in the progression:
(log(xy) + log(xy^2) + ...) - log(x) = log(xy^n) - log(x^n)
Now, let's substitute x = 160 into the expression and simplify:
Difference = log(160y^n) - log(160^n)
= log((160y^n) / 160^n)
= log(y^n / (160^(n-1)))
To make the difference constant, the expression inside the logarithm should be independent of n. Let's analyze the term:
y^n / (160^(n-1))
Since we want the expression to be independent of n, the exponents of y and 160 should cancel each other out. For that to happen, y should be equal to 160.
By setting y = 160, the difference becomes:
Difference = log(160^n / 160^(n-1))
= log(160 / 160^(1-1))
= log(160 / 160^0)
= log(160)
Therefore, when x = 160, the difference between consecutive terms in the progression is log(160), which is a constant. Hence, the expression log(x) + log(xy) + log(xy^2) + ... forms an arithmetic progression.
Now moving to the second question:
To sum the series 2 + 6 + 18 + 54 ... up to the 30th term, we can observe that each term is obtained by multiplying the previous term by 3. This is a geometric progression with a common ratio of 3.
The formula to calculate the sum of a geometric progression is:
Sum = (first term * (1 - common ratio^n)) / (1 - common ratio)
In this case, the first term is 2 (a = 2) and the common ratio is 3 (r = 3). We want to calculate the sum up to the 30th term, so n = 30.
Sum = (2 * (1 - 3^30)) / (1 - 3)
= (2 * (1 - 243)) / (-2)
= (2 * (-242)) / (-2)
= 484
By using the logarithm, we can verify that this sum is less than 2 x 10^14:
log(484) < log(2 x 10^14)
2.685 < 14.301
Since 2.685 is indeed less than 14.301, we can verify that the sum of the series 2 + 6 + 18 + 54 ... up to the 30th term is less than 2 x 10^14.