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

- anonymous

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

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- schrodinger

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- JamesJ

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

- anonymous

x=1.225

- 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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## More answers

- 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

could you help me with another problem? \[9^{x-5} = 81^{3}\]

- anonymous

waiiit.. james j :/ hahah

- myininaya

\[\log(ab)=\log(a)+\log(b)\]

- myininaya

Do you recall this?

- anonymous

yes

- myininaya

\[\log(5)=\log(4(-4x+5))\]

- myininaya

so now we can set the insides equal

- myininaya

\[5=4(-4x+5)\]

- anonymous

THAT'S WHY I HAD IT MULTIPLYING. haha sweet (:

- sasogeek

##### 1 Attachment

- myininaya

\[5=-16x+20\]
I distributed the 4 here

- myininaya

Now I'm going to subtract 20 on both sides
\[5-20=-16x\]

- myininaya

\[-15=-16x => -16x=-15 \]

- myininaya

Now we divide both sides by -16
\[x=\frac{-15}{-16}\]

- myininaya

But a negative divided by a negative is positive

- myininaya

\[x=\frac{15}{16}\]

- sasogeek

what did i do wrong? my answer doesn't concur with yours myininaya... :|

- 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

i know my working is wrong, i don't see where i went wrong though...

- anonymous

myininaya could you help me with the other equation i put up there please?

- myininaya

@saso I can't read yours that well

- anonymous

sasogeek, you didn't just cancle out the log, you cancled the five as well, right?

- anonymous

thanks for ur advice james

- sasogeek

log5-log5=0

- myininaya

Taranicolee you can post if you wish

- anonymous

sasogeek you have to distribute the four before you subtract from in the parenthesis

- anonymous

9 x−5 =81 3

- anonymous

oh wait that's not it

- myininaya

by the way \[\log(4x+5) \neq \log(4x)+\log(5) \]

- anonymous

\[9^{x -5}=81^{3}\]

- sasogeek

i'm just noticing that, thanks for pointing it out. i forgot the log(ab)=log(a)+log(b) rule :)

- myininaya

\[81=9^2 \]
agree?

- anonymous

yes

- myininaya

\[9^{x-5}=(9^2)^3\]

- myininaya

so i just replaced 81 with 9 squared

- anonymous

ohh to get the bases the same so you can cancle out right?

- myininaya

now law of exponents tells us \[(x^a)^b=x^{a \cdot b}\]

- 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

\[9^{x-5}=9^6 =>x-5=6\]

- 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

do you recall the law of exponent i mentioned?

- anonymous

no :/

- myininaya

\[(x^a)^b=x^{a \cdot b} \]?

- anonymous

nope. i wasn't here for the teaching of this

- myininaya

yes if is it in the form you just multiply the exponents to simplify

- anonymous

*there

- myininaya

so for example if we have
\[(3^2)^3=3^{2 \cdot 3} =3^6 \]

- myininaya

or if we have
\[(43^4)^3=43^{4 \cdot 3} =43^{12}\]

- 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

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

2^4 and 2^3

- myininaya

yes! :)

- myininaya

\[(2^4)^x=2^3\]
so this is our equation

- anonymous

cancle bases, 4x=3 divide by 4 on both sides?

- myininaya

perfect! :)

- anonymous

YAYYYY. thank you !!!!

- myininaya

You're welcome.

- myininaya

Thank you for being eager to learn this stuff.

- anonymous

((:

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