${a^{m{{\log }_a}n}} = $
${a^{mn}}$
${m^n}$
${n^m}$
એકપણ નહીં
(c) ${a^{m{{\log }_a}n}} = {a^{{{\log }_a}{n^m}}} = {n^m}$.
સમીકરણ ${9^x} – {2^{x + {1 \over 2}}} = {2^{x + {3 \over 2}}} – {3^{2x – 1}}$ નો ઉકેલ મેળવો.
$\sqrt {[12 – \sqrt {(68 + 48\sqrt 2 )} ]} = $
જો ${\left( {{2 \over 3}} \right)^{x + 2}} = {\left( {{3 \over 2}} \right)^{2 – 2x}},$ તો $x =$
જો $x \ne 0 $ તો ${\left( {{{{x^l}} \over {{x^m}}}} \right)^{({l^2} + lm + {m^2})}}$${\left( {{{{x^m}} \over {{x^n}}}} \right)^{({m^2} + nm + {n^2})}}{\left( {{{{x^n}} \over {{x^l}}}} \right)^{({n^2} + nl + {l^2})}}=$
${({x^5})^{1/3}}{(16{x^3})^{2/3}}$${\left( {{1 \over 4}{x^{4/9}}} \right)^{ – 3/2}} = $
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