Question

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

Answer 1

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

Answer 2

x=1.225

Answer 3

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)

Answer 4

@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. :-)

Answer 5

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

Answer 6

waiiit.. james j :/ hahah

Answer 7

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

Answer 8

Do you recall this?

Answer 9

yes

Answer 10

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

Answer 11

so now we can set the insides equal

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

But a negative divided by a negative is positive

Answer 19

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

Answer 20

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

Answer 21

i think that's right though because when i tried to solve it on my own i got one, and 15/16 .9375

Answer 22

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

Answer 23

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

Answer 24

@saso I can't read yours that well

Answer 25

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

Answer 26

thanks for ur advice james

Answer 27

log5-log5=0

Answer 28

Taranicolee you can post if you wish

Answer 29

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

Answer 30

9 x−5 =81 3

Answer 31

oh wait that's not it

Answer 32

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

Answer 33

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

Answer 34

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

Answer 35

\[81=9^2 \]
agree?

Answer 36

yes

Answer 37

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

Answer 38

so i just replaced 81 with 9 squared

Answer 39

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

Answer 40

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

Answer 41

if you get the same base on both sides then the exponents have to be equal in order for the equation to hold

Answer 42

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

Answer 43

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?

Answer 44

do you recall the law of exponent i mentioned?

Answer 45

no :/

Answer 46

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

Answer 47

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

Answer 48

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

Answer 49

*there

Answer 50

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

Answer 51

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

Answer 52

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?

Answer 53

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

Answer 54

2^4 and 2^3

Answer 55

yes! :)

Answer 56

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

Answer 57

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

Answer 58

perfect! :)

Answer 59

YAYYYY. thank you !!!!

Answer 60

You're welcome.

Answer 61

Thank you for being eager to learn this stuff.

Answer 62

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