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Mathematics 13 Online
Nicole:

http://prntscr.com/noh6hx

Nicole:

@Narad

Narad:

It is false, the answer is \[(1/2)^{3}=0.125\]

Nicole:

okay http://prntscr.com/noh9cy

Narad:

\[7* \log _{7}4 =7*\ln4/\ln7\approx5\]

Nicole:

okay http://prntscr.com/nohb4g

Narad:

The equation is \[y=2^{x+1}+3\] Option D

Nicole:

okay http://prntscr.com/nohdao

Narad:

The range is \[ y \in [2, +\infty )\]

Nicole:

Okay http://prntscr.com/nohe9p

Narad:

The money in the account will be \[=P*e ^{rn} = 3000*e ^{0.025*5}=3399 $\]

Nicole:

Got it http://prntscr.com/nohf7i

Narad:

\[2\log _{3}5+ \log _{3}2= \log _{3}25+\log _{3}= \log _{3}75\]

Narad:

there is 2 missing

Nicole:

wheres the 2 thats missing?

Narad:

\[=\log _{3}25*2=\log _{3}50\]

Nicole:

so its 50 not 75?

Narad:

yes

Nicole:

okay http://prntscr.com/nohh5h

Narad:

\[x ^{6}=100=10^{2}\] \[6\log _{10}x=\log10_{10}10^{2}=2\] \[\log _{10}x=2/6=1/3\]

Narad:

There is a 10 in excess

Nicole:

okay but the final answer is log10x=2/6=1/3?

Narad:

yes

Nicole:

okay http://prntscr.com/nohjhs

Narad:

\[-2\log _{8}(x+1)=-8\] \[\log _{8}(x+1)=2\] \[x+1=8^{2}=64\] \[x=63\]

Nicole:

okay http://prntscr.com/nohmcm

Narad:

There is a correction x=8^4-1=4095

Narad:

\[7\log _{9}(x+8)=7\] \[\log _{9}(x+8)=1\] \[x+8=9^{1}=9\] x=1

Nicole:

okay'' http://prntscr.com/nohnjm

Narad:

true

Nicole:

okay http://prntscr.com/noho2e

Narad:

True

Nicole:

okay http://prntscr.com/nohoy9

Narad:

\[\log _{2}(4*3) = \log _{2}12\]

Nicole:

Okay http://prntscr.com/nohpwk

Narad:

False

Nicole:

okay http://prntscr.com/nohpzj

Narad:

True

Nicole:

okay last one: http://prntscr.com/nohqnd

Narad:

true

Nicole:

okay thanks so much

Narad:

You are welcome

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