A community for students.
Here's the question you clicked on:
 0 viewing
LaronJaxon
 2 years ago
What is the solution to the equation 1/sqrt 8= 4^(m  3) ?
LaronJaxon
 2 years ago
What is the solution to the equation 1/sqrt 8= 4^(m  3) ?

This Question is Closed

terenzreignz
 2 years ago
Best ResponseYou've already chosen the best response.1\[\large \frac{1}{\sqrt{8}}=4^{m3}\]

terenzreignz
 2 years ago
Best ResponseYou've already chosen the best response.1One nifty way of doing this is to try to express both sides of the equation as exponentials, ie, one base, one exponent... preferably with the same base. So, let's start with \[\frac{1}{\sqrt{8}}\] 8 is just 2³, so let's put it that way...\[\large \frac{1}{\sqrt{2^{3}}}\] And remember that taking the square root means raising something to the 1/2 power, so...\[\huge \frac{1}{\left( 2^3 \right)^{\frac{1}{2}}}\] Using laws of exponents, you get

terenzreignz
 2 years ago
Best ResponseYou've already chosen the best response.1\[\huge \frac{1}{2^{\frac{3}{2}}}\]Now remember that \[\large a^{n}=\frac{1}{a^n}\] So eventually, we're left with \[\huge \frac{1}{\sqrt{8}} = 2^{\frac{3}{2}}\] Now on to the other side of the equation...

terenzreignz
 2 years ago
Best ResponseYou've already chosen the best response.1\[\large 4^{m3}\] But 4 = 2² So, we can write it as \[\huge (2^2)^{m3}\] Again, using laws of exponents, it is just equal to \[\huge 2^{2(m3)}=2^{2m  6}\]

terenzreignz
 2 years ago
Best ResponseYou've already chosen the best response.1So, your problem becomes... \[\huge 2^{\frac{3}{2}}=2^{2m  6}\] Which can only mean \[\large \frac{3}{2}=2m  6\]

terenzreignz
 2 years ago
Best ResponseYou've already chosen the best response.1And the rest, is history :D
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.