Question Number 74753 by necxxx last updated on 30/Nov/19
$${evaluate}\:\mathrm{5}^{\sqrt{\mathrm{log}\:\mathrm{7}_{\mathrm{5}} }} \:−\:\mathrm{7}^{\sqrt{\mathrm{log}\:\mathrm{5}_{\mathrm{7}} }} \\ $$
Commented by mr W last updated on 30/Nov/19
$${what}\:{is}\:\mathrm{log}\:\mathrm{7}_{\mathrm{5}} ?\:{the}\:{same}\:{as}\:\mathrm{log}_{\mathrm{5}} \:\mathrm{7}\:? \\ $$
Commented by Kunal12588 last updated on 30/Nov/19
$${define}\:{log}\:{a}_{{b}} \\ $$
Commented by turbo msup by abdo last updated on 30/Nov/19
$${perhaps}\:\:{log}\mathrm{7}_{\mathrm{5}} \:\:{mean}\:{log}_{\mathrm{5}} \left(\mathrm{7}\right)... \\ $$
Commented by necxxx last updated on 01/Dec/19
$${exactly},\:{thats}\:{what}\:{i}\:{meant}.\:{i}\:{only} \\ $$$${didnt}\:{represent}\:{it}\:{properly}. \\ $$
Answered by MJS last updated on 30/Nov/19
![we had this before, in correct syntax a^(√(log_a b )) =a^(√((ln b)/(ln a))) = [ln (a^(√((ln b)/(ln a))) )=(√((ln b)/(ln a)))×ln a =(√(ln a ×ln b))] =e^(√(ln a ×ln b)) ⇒ answer is 0](Q74789.png)
$$\mathrm{we}\:\mathrm{had}\:\mathrm{this}\:\mathrm{before},\:\mathrm{in}\:\mathrm{correct}\:\mathrm{syntax} \\ $$$${a}^{\sqrt{\mathrm{log}_{{a}} \:{b}\:}} ={a}^{\sqrt{\frac{\mathrm{ln}\:{b}}{\mathrm{ln}\:{a}}}} = \\ $$$$\:\:\:\:\:\left[\mathrm{ln}\:\left({a}^{\sqrt{\frac{\mathrm{ln}\:{b}}{\mathrm{ln}\:{a}}}} \right)=\sqrt{\frac{\mathrm{ln}\:{b}}{\mathrm{ln}\:{a}}}×\mathrm{ln}\:{a}\:=\sqrt{\mathrm{ln}\:{a}\:×\mathrm{ln}\:{b}}\right] \\ $$$$=\mathrm{e}^{\sqrt{\mathrm{ln}\:{a}\:×\mathrm{ln}\:{b}}} \\ $$$$\Rightarrow\:\mathrm{answer}\:\mathrm{is}\:\mathrm{0} \\ $$
Commented by necxxx last updated on 01/Dec/19
$${thank}\:{you}\:{so}\:{much}.{I}\:{do}\:{understand}\:{it} \\ $$$${now}. \\ $$