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solve log5=log4+log(-4x+5)

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first simplify the RHS (right hand side), what do you get?
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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Other answers:

@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. :-)
could you help me with another problem? \[9^{x-5} = 81^{3}\]
waiiit.. james j :/ hahah
Do you recall this?
so now we can set the insides equal
1 Attachment
\[5=-16x+20\] I distributed the 4 here
Now I'm going to subtract 20 on both sides \[5-20=-16x\]
\[-15=-16x => -16x=-15 \]
Now we divide both sides by -16 \[x=\frac{-15}{-16}\]
But a negative divided by a negative is positive
what did i do wrong? my answer doesn't concur with yours myininaya... :|
i think that's right though because when i tried to solve it on my own i got one, and 15/16 .9375
i know my working is wrong, i don't see where i went wrong though...
myininaya could you help me with the other equation i put up there please?
@saso I can't read yours that well
sasogeek, you didn't just cancle out the log, you cancled the five as well, right?
thanks for ur advice james
Taranicolee you can post if you wish
sasogeek you have to distribute the four before you subtract from in the parenthesis
9 x−5 =81 3
oh wait that's not it
by the way \[\log(4x+5) \neq \log(4x)+\log(5) \]
\[9^{x -5}=81^{3}\]
i'm just noticing that, thanks for pointing it out. i forgot the log(ab)=log(a)+log(b) rule :)
\[81=9^2 \] agree?
so i just replaced 81 with 9 squared
ohh to get the bases the same so you can cancle out right?
now law of exponents tells us \[(x^a)^b=x^{a \cdot b}\]
if you get the same base on both sides then the exponents have to be equal in order for the equation to hold
\[9^{x-5}=9^6 =>x-5=6\]
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?
do you recall the law of exponent i mentioned?
no :/
\[(x^a)^b=x^{a \cdot b} \]?
nope. i wasn't here for the teaching of this
yes if is it in the form you just multiply the exponents to simplify
so for example if we have \[(3^2)^3=3^{2 \cdot 3} =3^6 \]
or if we have \[(43^4)^3=43^{4 \cdot 3} =43^{12}\]
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?
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
2^4 and 2^3
yes! :)
\[(2^4)^x=2^3\] so this is our equation
cancle bases, 4x=3 divide by 4 on both sides?
perfect! :)
YAYYYY. thank you !!!!
You're welcome.
Thank you for being eager to learn this stuff.

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