Mathematics
19 Online
OpenStudy (anonymous):
Solve 2 log x = log 64
14 years ago
Join the QuestionCove community and study together with friends!
Sign Up
myininaya (myininaya):
Using \[\log(x^r)=r \log(x)\]
I rewrote the equation
\[\log (x^2)=\log (64)\]
Set insides equal
By the way make sure you don't keep the negative answer if you get one :)
14 years ago
OpenStudy (anonymous):
so it the answer 64?
14 years ago
myininaya (myininaya):
Nope
14 years ago
myininaya (myininaya):
\[x^2=64 \text{ when x=?}\]
14 years ago
OpenStudy (anonymous):
8 !!!
14 years ago
Join the QuestionCove community and study together with friends!
Sign Up
OpenStudy (anonymous):
2logx = log64
logx2 = log64
elogx2=elog64 ##taking exponential both sides##
x2=64
x=8
14 years ago
myininaya (myininaya):
yes sofia x=8 is right! :)
14 years ago
OpenStudy (anonymous):
can you help me with this one myininaya, Solve log4 x – 1 = 2
14 years ago
myininaya (myininaya):
\[\log_4(x-1)=2 ?\]
14 years ago
myininaya (myininaya):
or\[\log_4(x)-1=2?\]
14 years ago
Join the QuestionCove community and study together with friends!
Sign Up
myininaya (myininaya):
or neither?
14 years ago
OpenStudy (anonymous):
ayyy i don't know this onee!! :S
14 years ago
myininaya (myininaya):
\[\log(4x-1)=2 \text{ or } \log(4x)-1=2\]
14 years ago
myininaya (myininaya):
I'm just trying to figure out the equation
14 years ago
myininaya (myininaya):
Like is it any of the ones I mentioned?
14 years ago
Join the QuestionCove community and study together with friends!
Sign Up
myininaya (myininaya):
\[\log_4(x)-1=2\]
If it is this one which I sorta think is the right interpretation of what you wrote
The first step is to add 1 on both sides
14 years ago
myininaya (myininaya):
\[\log_4(x)=3\]
14 years ago
myininaya (myininaya):
Now rewrite as an exponential equation! :)
\[x=4^3\]
14 years ago
myininaya (myininaya):
Or I mean an exponential form
14 years ago
myininaya (myininaya):
in*
14 years ago
Join the QuestionCove community and study together with friends!
Sign Up
myininaya (myininaya):
Any questions?
14 years ago