anonymous
  • anonymous
solve log5=log4+log(-4x+5)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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JamesJ
  • JamesJ
first simplify the RHS (right hand side), what do you get?
anonymous
  • anonymous
x=1.225
anonymous
  • anonymous
well it will equal 5=4+(-4x+5) and you would solve from there, but OH WAIT. haha for some reason when i was trying to solve it i wrote 4 times (-4x+5)

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JamesJ
  • JamesJ
@crazzy: it's actually almost never helpful just to give an answer without explaining how you got the answer. But now that you've given the answer, you can now explain step by step how you got there. :-)
anonymous
  • anonymous
could you help me with another problem? \[9^{x-5} = 81^{3}\]
anonymous
  • anonymous
waiiit.. james j :/ hahah
myininaya
  • myininaya
\[\log(ab)=\log(a)+\log(b)\]
myininaya
  • myininaya
Do you recall this?
anonymous
  • anonymous
yes
myininaya
  • myininaya
\[\log(5)=\log(4(-4x+5))\]
myininaya
  • myininaya
so now we can set the insides equal
myininaya
  • myininaya
\[5=4(-4x+5)\]
anonymous
  • anonymous
THAT'S WHY I HAD IT MULTIPLYING. haha sweet (:
sasogeek
  • sasogeek
1 Attachment
myininaya
  • myininaya
\[5=-16x+20\] I distributed the 4 here
myininaya
  • myininaya
Now I'm going to subtract 20 on both sides \[5-20=-16x\]
myininaya
  • myininaya
\[-15=-16x => -16x=-15 \]
myininaya
  • myininaya
Now we divide both sides by -16 \[x=\frac{-15}{-16}\]
myininaya
  • myininaya
But a negative divided by a negative is positive
myininaya
  • myininaya
\[x=\frac{15}{16}\]
sasogeek
  • sasogeek
what did i do wrong? my answer doesn't concur with yours myininaya... :|
anonymous
  • anonymous
i think that's right though because when i tried to solve it on my own i got one, and 15/16 .9375
sasogeek
  • sasogeek
i know my working is wrong, i don't see where i went wrong though...
anonymous
  • anonymous
myininaya could you help me with the other equation i put up there please?
myininaya
  • myininaya
@saso I can't read yours that well
anonymous
  • anonymous
sasogeek, you didn't just cancle out the log, you cancled the five as well, right?
anonymous
  • anonymous
thanks for ur advice james
sasogeek
  • sasogeek
log5-log5=0
myininaya
  • myininaya
Taranicolee you can post if you wish
anonymous
  • anonymous
sasogeek you have to distribute the four before you subtract from in the parenthesis
anonymous
  • anonymous
9 x−5 =81 3
anonymous
  • anonymous
oh wait that's not it
myininaya
  • myininaya
by the way \[\log(4x+5) \neq \log(4x)+\log(5) \]
anonymous
  • anonymous
\[9^{x -5}=81^{3}\]
sasogeek
  • sasogeek
i'm just noticing that, thanks for pointing it out. i forgot the log(ab)=log(a)+log(b) rule :)
myininaya
  • myininaya
\[81=9^2 \] agree?
anonymous
  • anonymous
yes
myininaya
  • myininaya
\[9^{x-5}=(9^2)^3\]
myininaya
  • myininaya
so i just replaced 81 with 9 squared
anonymous
  • anonymous
ohh to get the bases the same so you can cancle out right?
myininaya
  • myininaya
now law of exponents tells us \[(x^a)^b=x^{a \cdot b}\]
myininaya
  • myininaya
if you get the same base on both sides then the exponents have to be equal in order for the equation to hold
myininaya
  • myininaya
\[9^{x-5}=9^6 =>x-5=6\]
anonymous
  • anonymous
you multiplied the exponents on the right side of the equation right ? then cancled out the bases and now to solve you just add five to 6?
myininaya
  • myininaya
do you recall the law of exponent i mentioned?
anonymous
  • anonymous
no :/
myininaya
  • myininaya
\[(x^a)^b=x^{a \cdot b} \]?
anonymous
  • anonymous
nope. i wasn't here for the teaching of this
myininaya
  • myininaya
yes if is it in the form you just multiply the exponents to simplify
anonymous
  • anonymous
*there
myininaya
  • myininaya
so for example if we have \[(3^2)^3=3^{2 \cdot 3} =3^6 \]
myininaya
  • myininaya
or if we have \[(43^4)^3=43^{4 \cdot 3} =43^{12}\]
anonymous
  • anonymous
oh sweet ! you're awesome. wanna move to italy and be my tutor?? hahah. if the equation were \[16^{x}=8\] not in that form, how would you solve?
myininaya
  • myininaya
I would love to go to Italy. write both sides with same base both 16 and 8 can be written as 2 to the power of something
anonymous
  • anonymous
2^4 and 2^3
myininaya
  • myininaya
yes! :)
myininaya
  • myininaya
\[(2^4)^x=2^3\] so this is our equation
anonymous
  • anonymous
cancle bases, 4x=3 divide by 4 on both sides?
myininaya
  • myininaya
perfect! :)
anonymous
  • anonymous
YAYYYY. thank you !!!!
myininaya
  • myininaya
You're welcome.
myininaya
  • myininaya
Thank you for being eager to learn this stuff.
anonymous
  • anonymous
((:

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