## 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)

1. triciaal

anything to the zero = 1

2. triciaal

confused

3. triciaal

|dw:1442925138712:dw|

4. triciaal

|dw:1442925314771:dw|

5. anonymous

6. anonymous

@mathmate ?

7. anonymous

@Loser66

8. mathmate

@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 non-zero base equals zero, therefore $$x=log_{72}1=0$$ It will be similarly be true for the second one, $$x=log_{70}1=0\ 9. anonymous what? 10. anonymous no its 7^2 and 7^0 11. anonymous @mathmate 12. anonymous $(7^{2})^{x} =1$ 13. anonymous $(7^{0})^{x}=1$ 14. mathmate you can rewrite \((7^2)^x=1$$ as $$49^x=1$$ and the above explanation still holds as long as the base is >0.

15. anonymous

so what about part b? @mathmate

16. anonymous

@mathmate

17. mathmate

For 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) .