Number of value/s of x satisfy given eqution& | vedclass

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(d) ${5^{x - 1}} + 5\,{(0.2)^{x - 2}} = 26$$ \Rightarrow $${5^{x - 1}} + 5\,.\,{\left( {{1 \over 5}} \right)^{x - 2}} = 26$

==>${5^{x - 1}} + {5

Vedclass · Accepted Answer

both (b) and (c)

Vedclass · Answer

(d) ${5^{x - 1}} + 5\,{(0.2)^{x - 2}} = 26$$ \Rightarrow $${5^{x - 1}} + 5\,.\,{\left( {{1 \over 5}} \right)^{x - 2}} = 26$

==>${5^{x - 1}} + {5^{3 - x}} = 26$$ \Rightarrow $${5^{x - 1}} + 25\,.\,{5^{ - (x - 1)}} - 26 = 0$

==>${5^{2(x - 1)}} - 26.\,{5^{(x - 1)}} + 25 = 0$

==>${5^{2(x - 1)}} - {5^{x - 1}} - {25.5^{x - 1}} + 25 = 0$

==>${5^{x - 1}}({5^{x - 1}} - 1) - 25({5^{x - 1}} - 1) = 0$

==>$({5^{x - 1}} - 25)({5^{x - 1}} - 1) = 0$ ==>$({5^{x - 1}} - {5^2})\,({5^{x - 1}} - {5^0}) = 0$

==> ${5^{x - 1}} = {5^2}$or ${5^{x - 1}} = {5^0}$$ \Rightarrow $$x = 3,\,1$.

Number of value/s of $x$ satisfy given eqution ${5^{x - 1}} + 5.{(0.2)^{x - 2}} = 26$.

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