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Mathematics 16 Online
OpenStudy (anonymous):

help with simplifying logs and exponential functions? one is: logbase5(10)+logbase25(10)?

OpenStudy (anonymous):

=logbase5(10)+((logbase5(10))/(logbase5(25))) =logbase5(10)+((logbase5(10))/(logbase5(5^2))) =logbase5(10)+((logbase5(10))/(2logbase5(5))) =logbase5(10)+((logbase5(10))/(2*1)) =1.5logbase5(10) let me kno if need further simplification

OpenStudy (anonymous):

the answer on this sheet says 3/(2log5)?

OpenStudy (anonymous):

=1.5 (log10/log5) =1.5(1/log5) =3/2(1/log5) 3/(2log5)

OpenStudy (anonymous):

ohhh okay! coudl you help me with another?

OpenStudy (anonymous):

ok, lets see if am able to solve

OpenStudy (anonymous):

okay. (e^x - e^-x)^2 / e^(2x)?

OpenStudy (anonymous):

=(e^2x-2e^xe^-x+e^-2x)/e^2x =(e^2x-2+e^-2x)/e^2x =(e^2x/e^2x)-(2/e^2x)+(e^-2x/e^2x) =1-2e^-2x+e^-4x

OpenStudy (anonymous):

does it correspond to the answer

OpenStudy (anonymous):

(1-e^-2x)^2 is the answer

OpenStudy (anonymous):

only one step further 1-2e^-2x+e^-4x=(1-e^-2x)^2

OpenStudy (anonymous):

ohhh okay. one more?

OpenStudy (anonymous):

sorry me going nxt time am reaallllllllllllllyyyyyyyyyyyyyyyyyy soooooooooooorrrrrrrrrrrrrrrrrrryyyyyyyy

OpenStudy (anonymous):

its okay! thanks!

OpenStudy (anonymous):

post it nxt time i come i answer if u wouldnt hav already obtain the answer

OpenStudy (anonymous):

ok!

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