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Question Number 140863 by mathdanisur last updated on 13/May/21
1+(√3^a ) = 2(√2^a )
$$\mathrm{1}+\sqrt{\mathrm{3}^{\boldsymbol{{a}}} }\:=\:\mathrm{2}\sqrt{\mathrm{2}^{\boldsymbol{{a}}} } \\ $$
Commented by mr W last updated on 13/May/21
a=0 or 2
$${a}=\mathrm{0}\:{or}\:\mathrm{2} \\ $$
Commented by mathdanisur last updated on 13/May/21
thanks dear Sir solutions please
$${thanks}\:{dear}\:{Sir}\:{solutions}\:{please} \\ $$
Commented by mr W last updated on 13/May/21
there is no formula to calculate the solution. in this special case you can “see” the solution is a=0 or 2, because they are obvious. if the question is 1+(√3^a ) = 3(√2^a ) then the solution is not obvious. i think there is no way to get the exact solution. through such questions one can learn nothing and one can teach nothing.
$${there}\:{is}\:{no}\:{formula}\:{to}\:{calculate}\:{the} \\ $$$${solution}.\:{in}\:{this}\:{special}\:{case}\:{you}\:{can} \\ $$$$“{see}''\:{the}\:{solution}\:{is}\:{a}=\mathrm{0}\:{or}\:\mathrm{2},\:{because} \\ $$$${they}\:{are}\:{obvious}. \\ $$$${if}\:{the}\:{question}\:{is}\:\mathrm{1}+\sqrt{\mathrm{3}^{\boldsymbol{{a}}} }\:=\:\mathrm{3}\sqrt{\mathrm{2}^{\boldsymbol{{a}}} }\:{then} \\ $$$${the}\:{solution}\:{is}\:{not}\:{obvious}.\:{i}\:{think} \\ $$$${there}\:{is}\:{no}\:{way}\:{to}\:{get}\:{the}\:{exact} \\ $$$${solution}. \\ $$$${through}\:{such}\:{questions}\:{one}\:{can} \\ $$$${learn}\:{nothing}\:{and}\:{one}\:{can}\:{teach} \\ $$$${nothing}. \\ $$
Commented by mathdanisur last updated on 13/May/21
thank you dear Sir★
$${thank}\:{you}\:{dear}\:{Sir}\bigstar \\ $$
Answered by ajfour last updated on 14/May/21
p^2 =3^a q^2 =2^a ⇒ 2ln p=aln 3 2ln q=aln 2 p=((3q)/2) 1+p=2q 1+3q=2q ⇒ q=−1 ⇒ 2^a =1 ⇒ a=0 p=−3 3^a =9 ⇒ a=2
$${p}^{\mathrm{2}} =\mathrm{3}^{{a}} \:\:\:\:{q}^{\mathrm{2}} =\mathrm{2}^{{a}} \\ $$$$\Rightarrow\:\:\mathrm{2ln}\:{p}={a}\mathrm{ln}\:\mathrm{3} \\ $$$$\:\:\:\:\:\:\:\mathrm{2ln}\:{q}={a}\mathrm{ln}\:\mathrm{2} \\ $$$$\:\:{p}=\frac{\mathrm{3}{q}}{\mathrm{2}} \\ $$$$\mathrm{1}+{p}=\mathrm{2}{q} \\ $$$$\mathrm{1}+\mathrm{3}{q}=\mathrm{2}{q} \\ $$$$\Rightarrow\:\:{q}=−\mathrm{1} \\ $$$$\:\:\:\Rightarrow\:\:\mathrm{2}^{{a}} =\mathrm{1}\:\:\Rightarrow\:\:{a}=\mathrm{0} \\ $$$${p}=−\mathrm{3} \\ $$$$\mathrm{3}^{{a}} =\mathrm{9}\:\:\:\Rightarrow\:\:{a}=\mathrm{2} \\ $$
Commented by prakash jain last updated on 14/May/21
How did you get p=((3q)/2)?
$$\mathrm{How}\:\mathrm{did}\:\mathrm{you}\:\mathrm{get}\:\mathrm{p}=\frac{\mathrm{3q}}{\mathrm{2}}? \\ $$
Commented by bramlexs22 last updated on 14/May/21
⇔ a = a ⇔ ((2ln p)/(ln 3)) = ((2ln q)/(ln 2)) ⇔ ln p = ((ln 3. ln q)/(ln 2)) ⇒ ln p = ((ln q^(ln 3) )/(ln 2)) ⇒ln p = ln q^(ln 3^(1/(ln 2)) ) ⇒ p = q^(ln 3^( (1/(ln 2))) )
$$\Leftrightarrow\:\mathrm{a}\:=\:\mathrm{a} \\ $$$$\Leftrightarrow\:\frac{\cancel{\mathrm{2}ln}\:\mathrm{p}}{\mathrm{ln}\:\mathrm{3}}\:=\:\frac{\cancel{\mathrm{2}ln}\:\mathrm{q}}{\mathrm{ln}\:\mathrm{2}} \\ $$$$\Leftrightarrow\:\mathrm{ln}\:\mathrm{p}\:=\:\frac{\mathrm{ln}\:\mathrm{3}.\:\mathrm{ln}\:\mathrm{q}}{\mathrm{ln}\:\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{ln}\:\mathrm{p}\:=\:\frac{\mathrm{ln}\:\mathrm{q}^{\mathrm{ln}\:\mathrm{3}} }{\mathrm{ln}\:\mathrm{2}} \\ $$$$\Rightarrow\mathrm{ln}\:\mathrm{p}\:=\:\mathrm{ln}\:\mathrm{q}^{\mathrm{ln}\:\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{ln}\:\mathrm{2}}} } \\ $$$$\Rightarrow\:\mathrm{p}\:=\:\mathrm{q}^{\mathrm{ln}\:\mathrm{3}^{\:\frac{\mathrm{1}}{\mathrm{ln}\:\mathrm{2}}} } \\ $$
Commented by prakash jain last updated on 14/May/21
It isn′t same as p=((3q)/2)
$$\mathrm{It}\:\mathrm{isn}'\mathrm{t}\:\mathrm{same}\:\mathrm{as}\:\mathrm{p}=\frac{\mathrm{3q}}{\mathrm{2}} \\ $$

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