What is the logarithmic equation for 3e3y-6 = 2x2-1?

[fr33pl4gu3]

3e^3y-6 = 2x²-1
ln3e^3y-6=ln2x²-lne
ln3e^3y-6=ln2x²/lne
3e^3y-6=2x²/e
3e^3y-6*e=2x²
e^3y-6=(2/3)x²
3y-6=ln((2/3)x²)
3y=ln((2/3)x²)+6
y=(1/3)ln((2/3)x²)+2

I wonder the answer is correct, the question is asking about the equation of y.

 

[HallsofIvy]

3e^3y-6 = 2x²-1
ln3e^3y-6=ln2x²-1

Be carefule with parentheses: you mean ln3e^3y-6=ln(2x²-1) because you are taking the log of the entire right side: 2x²- 1.

ln3e^3y-6=ln2x²-lne

You see? Now you are treating that "-1" as if it were outside the logarithm: it isn't.

ln3e^3y-6=ln2x²/lne

And even if it were, "ln a- ln b" is NOT "ln a/ln b" it is ln(a/b)

3e^3y-6=2x²/e

If you reverse what you did in the first step you should just come back to the first step! You don't because you have done it wrong.
ln(3e^{3y- 6})= ln(e3y- 6)+ ln 3= 3y- 6+ ln 3
3y- 6+ ln 3= ln(2x²- 1)

Solve that for y.

3e^3y-6*e=2x²
e^3y-6=(2/3)x²
3y-6=ln((2/3)x²)
3y=ln((2/3)x²)+6
y=(1/3)ln((2/3)x²)+2

I wonder the answer is correct, the question is asking about the equation of y.

No, make the corrections I suggested.

 

[fr33pl4gu3]

ln(3e3y- 6)= ln(e^{3y- 6})+ ln 3= 3y- 6+ ln 3
3y- 6+ ln 3= ln(2x²- 1)
3y-6 = ln((2x²-1)/3)
3y = ln ((2x²-1)/3) + 6
y=(1/3)ln((2x²-1)/3)+2

 

[rootX]

Yes, that's what I got

 

[fr33pl4gu3]

New Question:

log₈(9y + 14) = 6x⁶-11
9y+14 = 8^6x6-11

The next step is it by entering this step??

 

[snipez90]

If you're solving for y again then yes that is correct as a next step.

 

[fr33pl4gu3]

the next 2 step will be:

ln 9y + 14 = ln8^6x6-11
ln 9y + 14 = 6x⁶-11 ln8

 

[fr33pl4gu3]

log₈(9y + 14) = 6x⁶-11
9y+14 = 8^6x6-11
ln (9y + 14) = ln8^6x6-11
ln (9y + 14) = (6x⁶-11) ln8
ln 9y + ln 14 = (6x⁶-11)ln8
ln9y=(6x⁶-11)ln8 - ln 14
y=((6x⁶-11)ln8 - ln 14)/ln9

Solve the equation of y, i wonder if this is right?

 

[HallsofIvy]

ln(3e3y- 6)= ln(e^{3y- 6})+ ln 3= 3y- 6+ ln 3

This is NOT the equation you gave before! Before it was ln(3e^{3y- 6})

3y- 6+ ln 3= ln(2x²- 1)

And if the "-6" was NOT in the exponent this is wrong. Did you mean "ln(3e^{3y- 6}"?

3y-6 = ln((2x²-1)/3)
3y = ln ((2x²-1)/3) + 6
y=(1/3)ln((2x²-1)/3)+2

 

[tiny-tim]

Hi fr33pl4gu3!

ln (9y + 14) = (6x⁶-11) ln8
ln 9y + ln 14 = (6x⁶-11)ln8

No … ln (9y + 14) is not the same as ln 9y + ln 14, is it?

btw you should learn the very useful formula ln_ab = lnb/lna