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anonymous
 4 years ago
solve log5=log4+log(4x+5)
anonymous
 4 years ago
solve log5=log4+log(4x+5)

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JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.0first simplify the RHS (right hand side), what do you get?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0well 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)

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.0@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
 4 years ago
Best ResponseYou've already chosen the best response.0could you help me with another problem? \[9^{x5} = 81^{3}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0waiiit.. james j :/ hahah

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.3\[\log(ab)=\log(a)+\log(b)\]

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.3\[\log(5)=\log(4(4x+5))\]

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.3so now we can set the insides equal

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0THAT'S WHY I HAD IT MULTIPLYING. haha sweet (:

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.3\[5=16x+20\] I distributed the 4 here

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.3Now I'm going to subtract 20 on both sides \[520=16x\]

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.3\[15=16x => 16x=15 \]

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.3Now we divide both sides by 16 \[x=\frac{15}{16}\]

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.3But a negative divided by a negative is positive

sasogeek
 4 years ago
Best ResponseYou've already chosen the best response.0what did i do wrong? my answer doesn't concur with yours myininaya... :

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i think that's right though because when i tried to solve it on my own i got one, and 15/16 .9375

sasogeek
 4 years ago
Best ResponseYou've already chosen the best response.0i know my working is wrong, i don't see where i went wrong though...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0myininaya could you help me with the other equation i put up there please?

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.3@saso I can't read yours that well

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sasogeek, you didn't just cancle out the log, you cancled the five as well, right?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0thanks for ur advice james

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.3Taranicolee you can post if you wish

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sasogeek you have to distribute the four before you subtract from in the parenthesis

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh wait that's not it

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.3by the way \[\log(4x+5) \neq \log(4x)+\log(5) \]

sasogeek
 4 years ago
Best ResponseYou've already chosen the best response.0i'm just noticing that, thanks for pointing it out. i forgot the log(ab)=log(a)+log(b) rule :)

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.3so i just replaced 81 with 9 squared

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ohh to get the bases the same so you can cancle out right?

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.3now law of exponents tells us \[(x^a)^b=x^{a \cdot b}\]

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.3if you get the same base on both sides then the exponents have to be equal in order for the equation to hold

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.3\[9^{x5}=9^6 =>x5=6\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you 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
 4 years ago
Best ResponseYou've already chosen the best response.3do you recall the law of exponent i mentioned?

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.3\[(x^a)^b=x^{a \cdot b} \]?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0nope. i wasn't here for the teaching of this

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.3yes if is it in the form you just multiply the exponents to simplify

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.3so for example if we have \[(3^2)^3=3^{2 \cdot 3} =3^6 \]

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.3or if we have \[(43^4)^3=43^{4 \cdot 3} =43^{12}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh 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
 4 years ago
Best ResponseYou've already chosen the best response.3I 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

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.3\[(2^4)^x=2^3\] so this is our equation

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0cancle bases, 4x=3 divide by 4 on both sides?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0YAYYYY. thank you !!!!

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.3Thank you for being eager to learn this stuff.
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