please help me in logarithm

[ethanjonas]

2log(3x-2) - log(x+1) = 1 + log2 (all are in base 10) , solve for x
this is my sol:
2(3x-2) / (x+1) = 1 + 2
== 6x-4 = 3(x+1)
== 6x-4 = 3x+3
== 6x-3x = 3 + 4
== 3x = 7
== x = 7/3

my book says that x is equal to 4!
how can that be?
thanks to those who can help! ^+^

 

[tkhunny]

2log(3x-2) = (3x-2)^2 not [2(3x-2)]

How did you get from "log(2)" to just "(2)"?

 

[soroban]

Hello, ethanjonas!

You had some strange steps in there . . .

\(\displaystyle \text{Solve for }x\!:\;\;2\log(3x-2) - \log(x+1) \:=\: 1 + \log(2)\;\;\text{(all base 10)}\)

I would solve it like this . . .


\(\displaystyle 2\log(3x-2) - \log(x+1) \;=\;1 + \log(2)\)

. . . . . . . . .\(\displaystyle \log(3x-2)^2 \;=\;1 + \log(2) + \log(x+1)\) .**

. . . . . . . . .\(\displaystyle \log(3x-2)^2 \;=\;\log(10) + \log(2) + \log(x+1)\)

. . . . . . . . .\(\displaystyle \log(3x-2)^2 \;=\;\log[20(x+1)]\)

\(\displaystyle \text{"Un-log":}\) . . . \(\displaystyle (3x-2)^2 \;=\;20(x+1)\)

. . . . . . . \(\displaystyle 9x^2 - 12x + 4 \;=\;20x + 20\)

. . . . . . \(\displaystyle 9x^2 - 32x - 16 \;=\;0\)

\(\displaystyle \text{Factor:}\). \(\displaystyle (x-4)(9x+4) \;=\;0\)


\(\displaystyle \text{Therefore: }\:x \,=\,4,\;\rlap{///////}x\,=\,-\tfrac{4}{9}\)


~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

**

I do this to avoid fractions . . . yeah, why not?

 

[ethanjonas]

2log(3x-2) = (3x-2)^2 not [2(3x-2)]

How did you get from "log(2)" to just "(2)"?

hi tkhunny!
i divided the whole expression with log, so i canceled it.
i found it in the internet.
by the way, i guess your right, it should be (3x-2)^2
thanks a lot!

 

[ethanjonas]

Hello, ethanjonas!

You had some strange steps in there . . .

\(\displaystyle \text{Solve for }x\!:\;\;2\log(3x-2) - \log(x+1) \:=\: 1 + \log(2)\;\;\text{(all base 10)}\)

I would solve it like this . . .


\(\displaystyle 2\log(3x-2) - \log(x+1) \;=\;1 + \log(2)\)

. . . . . . . . .\(\displaystyle \log(3x-2)^2 \;=\;1 + \log(2) + \log(x+1)\) .**

. . . . . . . . .\(\displaystyle \log(3x-2)^2 \;=\;\log(10) + \log(2) + \log(x+1)\)

. . . . . . . . .\(\displaystyle \log(3x-2)^2 \;=\;\log[20(x+1)]\)

\(\displaystyle \text{"Un-log":}\) . . . \(\displaystyle (3x-2)^2 \;=\;20(x+1)\)

. . . . . . . \(\displaystyle 9x^2 - 12x + 4 \;=\;20x + 20\)

. . . . . . \(\displaystyle 9x^2 - 32x - 16 \;=\;0\)

\(\displaystyle \text{Factor:}\). \(\displaystyle (x-4)(9x+4) \;=\;0\)


\(\displaystyle \text{Therefore: }\:x \,=\,4,\;\rlap{///////}x\,=\,-\tfrac{4}{9}\)


~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

**

I do this to avoid fractions . . . yeah, why not?

hi sir soroban!
hmmm, i think you got it right, but i don't get how 1 became log 10 if its a constant 1?
is there a property that says that a constant number can be converted into a log form?
just wondering?!
thanks a lot. ^_^

 

[ethanjonas]

i would like to grab this opportunity to ask you again of how to solve for x in this ln function:
ln(x+1)+ln(x-2) = ln4, the answer in the book: x = 3
my sol:
exponentiate expression: e^ln(x+1) + e^ln(x-2) = ln4
== x+1+x-2 = 4
== 2x = 5
== x = 5/2

what might be the problem again? thanks!

 

[tkhunny]

2log(3x-2) = (3x-2)^2 not [2(3x-2)]

How did you get from "log(2)" to just "(2)"?

hi tkhunny!
i divided the whole expression with log, so i canceled it.
i found it in the internet.
by the way, i guess your right, it should be (3x-2)^2
thanks a lot!

I was afraid of that. Please, never do that again. "log" means nothing unless you are in the forrestry business. log(x), now THAT means something.

Let's see: log(10) = 1. dividing by "log", I get 10 = 1/log. This should look very odd to you. It is meaningless.

Try it a little more abstractly. f(x) = 3x-4 Are you tempted to divide by 'f'? I hope not.

 

[lookagain]

"log" means nothing unless you are inthe forrestry business.

\(\displaystyle The \ antiderivative \ of \ \ \frac{1}{\text{cabin}} \ \ is \ \ \log \ {\text{cabin plus sea.}}\)

 

[soroban]

Hello, ethanjonas!

More strange steps!

\(\displaystyle \ln(x+1)+\ln(x-2) \:=\: \ln4\)


Exponentiate expression:

. . \(\displaystyle e^{\ln(x+1)} + e^{\ln(x-2)} \:=\: \ln4\) . No!

. . \(\displaystyle x+1+x-2 \:=\: 4\) . No-no-no!

\(\displaystyle \text{Try using the rules that are recommended on }this\text{ planet . . .}\)


. . . . . .\(\displaystyle \underbrace{\ln(x+1) + \ln(x-2)} \;=\;\ln 4\)

\(\displaystyle \text{"Un-log": }\;\ln[(x+1)(x-2)] \;=\;\ln 4\)

. . . . . . . . . .\(\displaystyle (x+1)(x-2) \;=\;4\)

. . . . . . . . . . . . \(\displaystyle x^2 - x - 2 \;=\;4\)

. . . . . . . . . . . . \(\displaystyle x^2 - x - 6 \;=\;0\)

\(\displaystyle \text{Factor: }\qquad\;\;(x-3)(x+2) \;=\;0\)


. . \(\displaystyle \text{Answer: }\;x = 3,\;\rlap{\,//////}x = -2\)

 

[ethanjonas]

2log(3x-2) = (3x-2)^2 not [2(3x-2)]

How did you get from "log(2)" to just "(2)"?

hi tkhunny!
i divided the whole expression with log, so i canceled it.
i found it in the internet.
by the way, i guess your right, it should be (3x-2)^2
thanks a lot!

I was afraid of that. Please, never do that again. "log" means nothing unless you are in the forrestry business. log(x), now THAT means something.

Let's see: log(10) = 1. dividing by "log", I get 10 = 1/log. This should look very odd to you. It is meaningless.

Try it a little more abstractly. f(x) = 3x-4 Are you tempted to divide by 'f'? I hope not.

haha. yeah, i actually came to that 1/log issue.
thanks for the advice!

 

[ethanjonas]

Hello, ethanjonas!

More strange steps!

\(\displaystyle \ln(x+1)+\ln(x-2) \:=\: \ln4\)


Exponentiate expression:

. . \(\displaystyle e^{\ln(x+1)} + e^{\ln(x-2)} \:=\: \ln4\) . No!

. . \(\displaystyle x+1+x-2 \:=\: 4\) . No-no-no!

\(\displaystyle \text{Try using the rules that are recommended on }this\text{ planet . . .}\)


. . . . . .\(\displaystyle \underbrace{\ln(x+1) + \ln(x-2)} \;=\;\ln 4\)

\(\displaystyle \text{"Un-log": }\;\ln[(x+1)(x-2)] \;=\;\ln 4\)

. . . . . . . . . .\(\displaystyle (x+1)(x-2) \;=\;4\)

. . . . . . . . . . . . \(\displaystyle x^2 - x - 2 \;=\;4\)

. . . . . . . . . . . . \(\displaystyle x^2 - x - 6 \;=\;0\)

\(\displaystyle \text{Factor: }\qquad\;\;(x-3)(x+2) \;=\;0\)


. . \(\displaystyle \text{Answer: }\;x = 3,\;\rlap{\,//////}x = -2\)

oh i see.
hmmm, now its coming to my senses!
we have just learned the basics in logarithmic functions that's why i don't know how to solve that yet.
thanks a lot for the help!
i think i will have an a+ on our quiz tomorrow!
))