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

x=1.225

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

waiiit.. james j :/ hahah

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

Do you recall this?

yes

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

so now we can set the insides equal

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

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

\[5=-16x+20\]
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

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

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?

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

thanks for ur advice james

log5-log5=0

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?

yes

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

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

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

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

*there

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

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

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.

((: