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 January 26th, 2016, 09:42 AM #1 Senior Member     Joined: Jan 2012 Posts: 140 Thanks: 2 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.
 January 26th, 2016, 10:57 AM #2 Math Team     Joined: Jul 2011 From: Texas Posts: 3,002 Thanks: 1587 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 from happy21 and Country Boy
January 27th, 2016, 05:06 AM   #3
Senior Member

Joined: Jan 2012

Posts: 140
Thanks: 2

Quote:
 Originally Posted by skeeter 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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