What missing exponent would make this true?? (25)(√6:(25))=5^(x)
What is the colon for?
sorry, mathway is gay and it didnt help and thats what is does for ^4sqrt
Oh, gotcha. I think putting a space between the ^4 and the sqrt will get rid of that problem. Does it look like: \[\frac{25}{(\sqrt{6})^4\sqrt{25}}=5^x\]
\[25\sqrt[4]{25}=5^{x}\]
i learned something new on here, thats what it looks like
Oh, gotcha 25 is the same as 5 squared, and the fourth root of 25 is the same as 25 to the 1/4 power. \[(5^2)(25^{\frac{1}{4}})\]Substitute 25 = 5 squred again. \[(5^2)((5^2)^{\frac{1}{4}})\]Simplify. \[(5^2)(5^{\frac{1}{2}})\]\[5^{2+\frac{1}{2}}\]\[5^{\frac{3}{2}}\]
I have to solve for x, and i have a choice, and what you put up there is not a choice :\
choices are : 7/2 , 7/6 , 1/3 , 7/3
Oh, shoot. \[5^{2+\frac{1}{2}}=5^{5/2}\] It should be 5/2, not 3/2 Was there a 6 somewhere in there? I saw one in the original equation, but not in the second time you posted.
Ah look how stupid i am, its not your fault, its mine yes there is a six not a four\[25\sqrt[6]{25}=5^{x}\]
Oh, okay. I was worried, because I kept getting 5/2. Are you okay solving it? It's the same steps.
yeah i can try lol
Okay, I'll stick around just in case you need help. The solution is one of those choices.
yeah i need help XD.
Sorry, im not good wit this
No worries, how far did you get?
to be honest, i dont understand this and mathway abd wolframalpha has been my thing on this, I have no clue how to do this
Okay. So are you okay with \[25=5^2; \sqrt[6]{25}=25^{1/6}\]
yes
Okay, so rewriting the equation, we get \[(5^2)(25^{1/6})=5^x\] So the first step is to get rid of any radical signs.
k
By looking at the equation, we notice that 5 is the base for two of the exponentials and 25 is the base for the other one. It would be nice to make them all have the same base. \[(5^2)(5^2)^{1/6}=5^x\] So, second step: Try to make all the bases the same.
How are you doing so far?
sorry, i got to go. summer school out, maybe we do this later, i just guessed on my test and failed
Alright. Let me know if you ever want help. Take care.
Join our real-time social learning platform and learn together with your friends!