Problem of the Week

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Any help would be greatly appreciated. Thanks in advance.

What is the smallest possible degree measure for which (16^(sin^2X))(4^(2sinX))(2)=1

sin^2X is saying sine squared X
2sinX is saying 2 times the sine of X

Thanks
Please show all work
 
Work with logs to find the solution

=(16^(sin^2X))

rewrite this as

= log 16 etc

= log (2^3)

=3log2

do this for each of the parts, working to a common part of log 2
(note log (-1/2) = log 1 - log 2 = 0 - log2)
 
I appreciate the help, but logs arent my area of expertise. Could you possibly solve this equation, and write out the steps for me in more detail?
 
I could do all the steps but it will help you learn if you work with me and show what you have done - I will edit and assist your attempts.

(16^(sin^2X))(4^(2sinX))(2)=1


(16^(sin^2X))(4^(2sinX))= 1 /2

log a + log b = log (1/2)

whera a =(16^(sin^2X))
and b= (4^(2sinX))

Now use the idea of 16 = 2^3, and 4 = 2^2


and also recall that log (f^2) = 2 log f

now try and show where you get to.
 
I'm still not sure I get where your going next. I mean, I can follow your steps up to this point, but I dont know where to go from that. Could you possibly show me a little more, and then I'll spend some time working on the rest?
 
Where do you go from this:


log 16^(sin^2X) + log 4^2sinX = log 1/2
 
log 16^(sin^2X) + log 4^2sinX = log 1/2

log 2^(4 sin^2 x) + log 2^(2 sin x) = log 1 - log 2


as 2^4 = 16
and 2^2 = 4
 
yes you are correct, how could it be-sorry for the typo.

log 4^2sinx= log 2 ^(4 sin x)
 
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