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Difference between revisions of "1976 AHSME Problems/Problem 20"

(Created page with "Let <math>a,~b</math>, and <math>x</math> be positive real numbers distinct from one. Then <math>4(\log_ax)^2+3(\log_bx)^2=8(\log_ax)(\log_bx)</math> <math>\textbf{(A) }\text...")
 
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Because <math>\log_mn = \dfrac{\log{n}}{\log{m}}</math>, <math>4(\log_ax)^2+3(\log_bx)^2 = \dfrac{4(\logx)^2}{(\loga)^2}+\dfrac{3(\logx)^2}{(\logb)^2} = \dfrac{(\logx)^2(4(\loga)^2+3(\logb)^2)}{(\loga\logb)^2}</math>.
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Because <math>\log_{m} n = \dfrac{\log n}{\log m}</math>, <math>4(\log_{a} x)^2+3(\log_{b} x)^2 = </math> <math> \dfrac{4(\log x)^2}{(\log a)^2}+\dfrac{3(\log x)^2}{(\log b)^2} =\dfrac{(\log x)^2(4(\log a)^2+3(\log b)^2)}{(\log a \log b)^2}</math>.

Revision as of 04:32, 27 November 2021

Let $a,~b$, and $x$ be positive real numbers distinct from one. Then $4(\log_ax)^2+3(\log_bx)^2=8(\log_ax)(\log_bx)$

$\textbf{(A) }\text{for all values of }a,~b,\text{ and }x\qquad\\ \textbf{(B) }\text{if and only if }a=b^2\qquad\\ \textbf{(C) }\text{if and only if }b=a^2\qquad\\ \textbf{(D) }\text{if and only if }x=ab\qquad\\ \textbf{(E) }\text{for none of these}$


Solution

Because $\log_{m} n = \dfrac{\log n}{\log m}$, $4(\log_{a} x)^2+3(\log_{b} x)^2 =$ $\dfrac{4(\log x)^2}{(\log a)^2}+\dfrac{3(\log x)^2}{(\log b)^2} =\dfrac{(\log x)^2(4(\log a)^2+3(\log b)^2)}{(\log a \log b)^2}$.

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