Need Help with Question

[chowmeinness]

25^x=8

5^(x-1)=?

how do you solve this problem?

Much appreciated!

 

[biffboy]

25^x=8

5^(x-1)=?

how do you solve this problem?

Much appreciated!

25^x=8 (5^2)^x=8 5^2x=8 Taking square roots 5^x=root8 5^(x-1)=5^x/5=(root8)/5=0.566

 

[Deleted member 4993]

25^x=8

5^(x-1)=?

how do you solve this problem?

Much appreciated!

Same as post above - but displayed in LaTex

\(\displaystyle 25^x \ = \ 8\)

\(\displaystyle 5^x \ = \ \sqrt{8}\)

\(\displaystyle \dfrac{5^x}{5} \ = \ \dfrac{\sqrt{8}}{5}\)

\(\displaystyle 5^{x \ - \ 1} \ = \ \dfrac{\sqrt{8}}{5}\)

 

[chowmeinness]

thank you for everyone who helped solve this problem. I know how to do it now.

 

[mmm4444bot]

I know how to do it now.

Cool.

What will happen when you get a different it? Like this:

24^x = 8

5^(x - 1) = ?

 

[Deleted member 4993]

Can you "LaTex" in red?

I let Mark borrow my bottle of red-ink.

 

[lookagain]

25^x = 8

(5^2)^x = 8

5^2x = 8 **
No, that whole exponent needs grouping symbols around it:
5^(2x) = 8. Otherwise it's equivalent to (5^2)(x) = 8.

Taking square roots:

5^x = root8 . . . .

It would be \(\displaystyle 5^x = \pm\sqrt{8}, \) from which
you would have to discard the negative value.

5^(x-1) = 5^x/5
With your reasoning from **, this would be equivalent to 5^(x - 1) = 5^(x/5).
That is, you can't have it both ways.


(root8)/5 ~ 0.566

biffboy,

use separate lines please. You are showing that one equation
leads to the next equation. You are not to be putting equals signs
between equations. If anything, you could put a "==>" symbol
between them for implication.