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જો $1,\,{\log _9}\,\left( {{3^{1 - x}}\, + \,2} \right),\,\,{\log _3}\,\left( {{{4.3}^x}\, - \,1} \right)$
સમાંતર શ્રેણીમાં ,હોય તો ${\text{x = }}........$
$log_3\ 4$
$1 - log_3\ 4$
$1 - log_4\ 3$
$log_4\ 3$
Solution
અહીં $,\,\,1,\,{\log _9}\,\left( {{3^{1 – x}}\, + \,2} \right),\,{\log _3}\,\left( {{{4.3}^x}\, – \,1} \right)$ સમાંતર શ્રેણીમાં છે.
$\therefore \,\,\,2\,{\log _9}\,\left( {{3^{1 – x}}\, + \,2} \right)\,\, = \,\,1\, + \,{\log _3}\,\left( {{{4.3}^x}\, – \,1} \right)$
$\,\therefore \,\,\,\,\,\frac{2}{2}\,{\log _3}\,\left( {{3^{1 – x}}\, + \,2} \right)\, = \,\,{\log _3}^3\, + \,{\log _3}\,\left( {{{4.3}^x}\, – \,1} \right)$
$\therefore \,\,\,{\log _9}\,\left( {{3^{1 – x}}\, + \,2} \right)\,\, = \,\,{\log _3}\,3\,\left( {{{4.3}^x}\, – \,1} \right)\,\,$
$\therefore \,\,\,\,\,{3^{1\, – \,x}}\, + \,2\,\, = \,\,3\,\,\left( {4\,{{.3}^x}\, – \,1} \right)$
$\therefore \,\,\frac{3}{{{3^x}}}\, + \,2\,\, = \,\,{12.3^x}\, – \,3$
$12y^2 – 5y – 3 = 0$ જ્યાં $y = 3^x$
$12y^2 – 9y + 4y – 3 = 0$
$(4y – 3) (y +1) = 0$
$ y = 3/4$ અથવા $y = -1$
$\therefore \,\,y\,\, = \,\,\frac{3}{4}\,\,\left[ {\because \,y\,\, = \,\,{3^x}\,\, \ne \, – \,1} \right]$
$\,\therefore \,\,{3^x}\,\, = \,\,\frac{3}{4}\,\,\,$
$\,\therefore \,\,x\,\, = \,\,{\log _3}\,\frac{3}{4}\,\, = \,\,{\log _3}\,3\, – \,{\log _3}\,4\, = \,1\, – \,{\log _3}\,4$
