SOLUTION: In a geometric progression series find the value of r if logr+logr^2+logr^4+logr^8+...+logr^32=63

Question 1134866: In a geometric progression series find the value of r if logr+logr^2+logr^4+logr^8+...+logr^32=63
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
it is unfortunate that they used the designation of r in the log because that gets it confused with the rate in the formula of Sn = A1 * (1 - r^n) / (1 - r).

that formula represents the sum of a geometric series.

i changed the r to an x and the problem now reads.

log(x) + log(x^2) + log(x^4) + log(x^8) + ... log(x^32) = 63.

the missing term between log(x^8) and log(x^32) looks like it needs to be log(x^16).

the series then becomes log(x) + log(x^2) + log(x^4) + log(x^8) + log(x^16) + log(x^32) = 63.

since log(x^n) = n * log(x), that series can now be shown as:

log(x) + 2 * log(x) + 4 * log(x) + 8 * log(x) + 16 * log(x) + 32 * log(x) = 63

the common ratio is 2, because each term in the series doubles in size.

there are 6 terms in the series.

the first terms is log(x).

the formula for the sum of a geometric series is Sn = A1 (1 - r^n) / (1 - r)

when A1 = log(x) and n = 6 and r = 2, this formula becomes:

Sn = log(x) * (1 - 2^6) / (1 - 2).

since Sn = 63, this formula becomes 63 = log(x) * (1 - 2^6) / (1 - 2)

simplify this formula to get 63 = log(x) * -63 / -1.

simplify to get 63 = log(x) * 63.

solve for log(x) to get log(x) = 1

this is true if and only if 10^1 = x.

this makes x = 10.

when x = 10 and A1 = log(x) and r = 2 and n = 6 and Sn = 63, Sn = A1 * (1 - r^n) / (1 - r) becomes 63 = log(10) * (1 - 2^6) / (1 - 2) which becomes 63 = 1 * -63 / -1 which becomes 63 = 63 which is true.

your solution appears to be be x = 10.

going back to the original series where r was used in place of x, you would get:

log(10) + log(10^2) + log(10^4) + log(10^8) + log(1-^16) + log(10^32) = 63

you can use your calculator to confirm that this last equation is true.

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