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January 26th, 2016, 09:42 AM   #1
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A queston on logarithms

Hello,

Can one help me solve one logarithms' question:

Given aand b are positive numbers satisfying $\displaystyle 4(\log_{10}a)^{2}+(\log_{2}b)^{2}=1$, then which of the following statement(s) is/are correct?

(A) Greatest and least possible values of 'a' are reciprocal of each other.
(B) Greatest and least possible values of 'b' are reciprocal of each other.
(C) Greatest value of 'a' is the square of the largest value of 'b'.
(D) Least value of 'b' is the square of the least value of 'a'.

Thanks.
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January 26th, 2016, 10:57 AM   #2
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let $x=\log_{10}{a}$

$y = \log_2{b}$

$4x^2+y^2 = 1$, an ellipse such that $-\dfrac{1}{2} \le x \le \dfrac{1}{2}$ and $-1 \le y \le 1$

$-\dfrac{1}{2} \le \log_{10}{a} \le \dfrac{1}{2}$

$-1 \le \log_2{b} \le 1$

... enough information for you to continue?
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January 27th, 2016, 05:06 AM   #3
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Quote:
Originally Posted by skeeter View Post
let $x=\log_{10}{a}$

$y = \log_2{b}$

$4x^2+y^2 = 1$, an ellipse such that $-\dfrac{1}{2} \le x \le \dfrac{1}{2}$ and $-1 \le y \le 1$

$-\dfrac{1}{2} \le \log_{10}{a} \le \dfrac{1}{2}$

$-1 \le \log_2{b} \le 1$

... enough information for you to continue?
Thanks. The problem is now solved.
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