## Spring98 one year ago Part A: If (2^6)^x = 1, what is the value of x? Explain your answer. Part B: If (5^0)^x = 1, what are the possible values of x? Explain your answer.

1. Spring98

@jdoe0001 Can you help me with this one too. I can't pas my quiz and if i don't get 80% or more i can't go on.

2. anonymous

$$\bf \textit{part A, multiply both sides by }\qquad \cfrac{1}{26} \\ \quad \\ \textit{part B, multiply both sides by }\qquad \cfrac{1}{50}$$ what does that give you?

3. Spring98

they both equal 1 @jdoe0001

4. Spring98

is that right?

5. anonymous

$$\bf (26)x=1\implies \cfrac{1}{\cancel{26}}\cdot (\cancel{26})x=1\cdot \cfrac{1}{26}\implies x=? \\ \quad \\ \quad \\ (50)x=1\implies \cfrac{1}{\cancel{50}}\cdot (\cancel{50})x=1\cdot \cfrac{1}{50}\implies x=?$$

6. Spring98

0.0384615384615385 is part A and 0.02for part B @jdoe0001

7. anonymous

$$\huge \circ.\circ$$

8. anonymous

notice, the number on the left cancelled out with the denominator

9. Spring98

but what happened with the exponents?

10. Spring98

@jdoe0001

11. Spring98

@zepdrix @jdoe0001 can you guys plz help me?

12. anonymous

woops, got a typo there $$\bf (26)x=1\implies \cfrac{1}{\cancel{26}}\cdot (\cancel{26})x=1\cdot \cfrac{1}{26}\implies \cfrac{1}{1}\cdot x=\cfrac{1}{26}\implies x=\cfrac{1}{26} \\ \quad \\ \quad \\ (50)x=1\implies \cfrac{1}{\cancel{50}}\cdot (\cancel{50})x=1\cdot \cfrac{1}{50}\implies \cfrac{1}{1}\cdot x=\cfrac{1}{50}\implies x=\cfrac{1}{50}$$

13. Spring98

i know but there is expopnents and i'm going to get this wrong.

14. anonymous

ohh...shoot.. is not what you originally posted though hmm.... need to dash in secs..... you may want to repost

15. zepdrix

$\large\rm (2^{6})^x=1$Applying a rule of exponents to the left side gives us$\large\rm 2^{6x}=1$Rewrite the right side as a power of 2. 2 to what power = 1?

16. Spring98

ok i will repost and i will tag you k @jdoe0001

17. Spring98

Is it 0? @zepdrix

18. zepdrix

Good. Anything to the zero power is 1. 2^0=1. Therefore we can write our equation:$\large\rm 2^{6x}=1$as$\large\rm 2^{6x}=2^0$Since the bases are the same, we can equate the exponents.

19. Spring98

how did the 2 end up a 0. and isn't the x supposed to be a 0

20. Spring98

@zepdrix

21. zepdrix

I didn't change anything about the left side. I rewrite 1 as 2^0.

22. zepdrix

Therefore, equating the exponents gives us $$\large\rm 6x=0$$ which leads to $$\large\rm x=0$$

23. zepdrix

For the other problem:$\large\rm (5^0)^x = \color{orangered}{1}$again, rewrite 1 as a power of 5,$\large\rm (5^0)^x = \color{orangered}{5^0}$Apply exponent rule to the left side,$\large\rm (5^{0x}) = \color{orangered}{5^0}$Which leads to$\large\rm 0x=0$And it looks like ANY value of x is going to solve this one.