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Let Sn(x) = loga1/2 x + loga1/3 x + loga1/11 x loga18 x + loga1/27 x + .... up to n terms, where a > 1. If S24(x) = 1093 and S12(2x) = 265, then value of a is equal to _________.
Correct answer is '16'. Can you explain this answer?
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Let Sn(x) = loga1/2 x +loga1/3 x +loga1/11 x loga18 x +loga1/27 x + .....
To find Sn(x), we need to find the product of all the terms mentioned up to n terms.
We have:
Sn(x) = loga(1/2) * loga(1/3) * loga(1/11) * loga(18) * loga(1/27) * ...
Notice that the terms in the product alternate between positive and negative exponents. We can rearrange the terms to group the positive exponents together and the negative exponents together:
Sn(x) = (loga(1/2) * loga(18) * ...) * (loga(1/3) * loga(1/11) * loga(1/27) * ...)
The first group of terms can be simplified by using the property of logarithms:
loga(1/2) * loga(18) * ... = loga((1/2) * 18 * ...)
Similarly, the second group of terms can be simplified:
loga(1/3) * loga(1/11) * loga(1/27) * ... = loga((1/3) * (1/11) * (1/27) * ...)
Now, we can rewrite Sn(x) as:
Sn(x) = loga((1/2) * 18 * ...) * loga((1/3) * (1/11) * (1/27) * ...)
The terms inside each logarithm can be further simplified:
Sn(x) = loga((9) * ...) * loga((1/33) * ...)
Now, we can express Sn(x) as a single logarithm by multiplying the terms inside the logarithms:
Sn(x) = loga((9) * ...) * loga((1/33) * ...)
= loga((9) * (1/33) * ...)
= loga(9/33 * ...)
Therefore, Sn(x) = loga(9/33 * ...)
We have:
Sn(x) = loga(1/2) * loga(1/3) * loga(1/11) * loga(18) * loga(1/27) * ...
Notice that the terms in the product alternate between positive and negative exponents. We can rearrange the terms to group the positive exponents together and the negative exponents together:
Sn(x) = (loga(1/2) * loga(18) * ...) * (loga(1/3) * loga(1/11) * loga(1/27) * ...)
The first group of terms can be simplified by using the property of logarithms:
loga(1/2) * loga(18) * ... = loga((1/2) * 18 * ...)
Similarly, the second group of terms can be simplified:
loga(1/3) * loga(1/11) * loga(1/27) * ... = loga((1/3) * (1/11) * (1/27) * ...)
Now, we can rewrite Sn(x) as:
Sn(x) = loga((1/2) * 18 * ...) * loga((1/3) * (1/11) * (1/27) * ...)
The terms inside each logarithm can be further simplified:
Sn(x) = loga((9) * ...) * loga((1/33) * ...)
Now, we can express Sn(x) as a single logarithm by multiplying the terms inside the logarithms:
Sn(x) = loga((9) * ...) * loga((1/33) * ...)
= loga((9) * (1/33) * ...)
= loga(9/33 * ...)
Therefore, Sn(x) = loga(9/33 * ...)
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Community Answer
Let Sn(x) = loga1/2 x +loga1/3 x +loga1/11 x loga18 x +loga1/27 x + .....
Sn(x) = (2 + 3 + 6 + 11 + 18 + 27 + ... + n terms) logax
Let S1 = 2 + 3 + 6 + 11 + 18 + 27 + ... + Tn
S1 = 2 + 3 + 6 + ... + Tn _____
Tn = 2 + 1 + 3 + 5 + ... + n terms
Tn = 2 + (n - 1)2
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Let Sn(x) = loga1/2 x +loga1/3 x +loga1/11 x loga18 x +loga1/27 x + ....up to n terms, where a > 1. If S24(x) = 1093 and S12(2x) = 265, then value of a is equal to _________.Correct answer is '16'. Can you explain this answer?
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