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anonymous
 one year ago
Part A: If (72)x = 1, what is the value of x? Explain your answer. (5 points)
Part B: If (70)x = 1, what are the possible values of x? Explain your answer. (5 points)
anonymous
 one year ago
Part A: If (72)x = 1, what is the value of x? Explain your answer. (5 points) Part B: If (70)x = 1, what are the possible values of x? Explain your answer. (5 points)

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triciaal
 one year ago
Best ResponseYou've already chosen the best response.1anything to the zero = 1

triciaal
 one year ago
Best ResponseYou've already chosen the best response.1dw:1442925138712:dw

triciaal
 one year ago
Best ResponseYou've already chosen the best response.1dw:1442925314771:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@mathmate please help and explain too?

mathmate
 one year ago
Best ResponseYou've already chosen the best response.1@Tricial has pretty much explain everything. If you'd like to have a look in a different way, here it is. I assume your equations are 1. \(72^x=1\) and 2. \(70^x=1\) We can start with the definition of logarithms: \(y = log_a ~x ~~ <=> ~~ a^y = x\) and the fact that \(log_a 1=0\) and \(log_a a=1\) since \(a^0=1\) and \(a^1=a\) for all \(a\in R~and~ a> 0\) Applying the definition of logarithms, \(72^x=1~~<=>~~ log_{72}1=x\) But log(1) to any nonzero base equals zero, therefore \(x=log_{72}1=0\) It will be similarly be true for the second one, \(x=log_{70}1=0\

mathmate
 one year ago
Best ResponseYou've already chosen the best response.1you can rewrite \((7^2)^x=1\) as \(49^x=1\) and the above explanation still holds as long as the base is >0.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so what about part b? @mathmate

mathmate
 one year ago
Best ResponseYou've already chosen the best response.1For part b, 7^0=1. Since 1 raised to any power, 1*1*1*1*1....*1 = 1, so there are many possible solutions for x. For example, 1^0=1, 1^1=1, 1^2=1, ... 1^2999=1. so x=any integer or rational number. For example, 1^(15/37)= (1^15) to the 37th root is still 1. When x is an irrational number, x cannot be evaluated using the definition of logarithms since logarithm to the base 1 is not defined. In that case, perhaps the use of limits will still define 1^x =1 (where x is irrational) .
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