0000:00:00:00:00Start19Number 19ExplanationIf aaa satisfies the equation log22x+log33x=log44x2\log_2 2x + \log_3 3x = \log_4 4x^2log22x+log33x=log44x2, then the value of loga3=....\log_a 3 = ....loga3=....−3-3−3333111−1-1−1000ExplanationRecall some logarithm concepts logab=1logba\log_a b = \frac{1}{\log_b a}logab=logba1logabc=logab+logac\log_a bc = \log_a b + \log_a clogabc=logab+logac And also logambn=nm⋅logab\log_{a^m} b^n = \frac{n}{m} \cdot \log_a blogambn=mn⋅logab. Let's simplify the logarithm form log22x+log33x=log44x2\log_2 2x + \log_3 3x = \log_4 4x^2log22x+log33x=log44x2log22+log2x+log33+log3x=log44+log4x2\log_2 2 + \log_2 x + \log_3 3 + \log_3 x = \log_4 4 + \log_4 x^2log22+log2x+log33+log3x=log44+log4x21+log2x+1+log3x=1+log22x21 + \log_2 x + 1 + \log_3 x = 1 + \log_{2^2} x^21+log2x+1+log3x=1+log22x21+log2x+log3x=22⋅log2x1 + \log_2 x + \log_3 x = \frac{2}{2} \cdot \log_2 x1+log2x+log3x=22⋅log2xlog3x=−1\log_3 x = -1log3x=−1logx3=1−1=−1\log_x 3 = \frac{1}{-1} = -1logx3=−11=−1 Therefore the solution is x=ax = ax=a so loga3=−1\log_a 3 = -1loga3=−1
19Number 19ExplanationIf aaa satisfies the equation log22x+log33x=log44x2\log_2 2x + \log_3 3x = \log_4 4x^2log22x+log33x=log44x2, then the value of loga3=....\log_a 3 = ....loga3=....−3-3−3333111−1-1−1000ExplanationRecall some logarithm concepts logab=1logba\log_a b = \frac{1}{\log_b a}logab=logba1logabc=logab+logac\log_a bc = \log_a b + \log_a clogabc=logab+logac And also logambn=nm⋅logab\log_{a^m} b^n = \frac{n}{m} \cdot \log_a blogambn=mn⋅logab. Let's simplify the logarithm form log22x+log33x=log44x2\log_2 2x + \log_3 3x = \log_4 4x^2log22x+log33x=log44x2log22+log2x+log33+log3x=log44+log4x2\log_2 2 + \log_2 x + \log_3 3 + \log_3 x = \log_4 4 + \log_4 x^2log22+log2x+log33+log3x=log44+log4x21+log2x+1+log3x=1+log22x21 + \log_2 x + 1 + \log_3 x = 1 + \log_{2^2} x^21+log2x+1+log3x=1+log22x21+log2x+log3x=22⋅log2x1 + \log_2 x + \log_3 x = \frac{2}{2} \cdot \log_2 x1+log2x+log3x=22⋅log2xlog3x=−1\log_3 x = -1log3x=−1logx3=1−1=−1\log_x 3 = \frac{1}{-1} = -1logx3=−11=−1 Therefore the solution is x=ax = ax=a so loga3=−1\log_a 3 = -1loga3=−1
