Oh, this is one of those times where you have to remember that identities work both ways. Here we go. log(3x+7) + log(x-2) = 1 There’s a clever saying, “Everyone knows 1+1=2, but I KNOW 2=1+1.” Indeed, solving this hinges on understanding that: log(m*n) = log(m) + log(n) Also means: log(m) + log(n) = log(m*n) Here we have the sum of two logs with the same base, therefore we can combine them like so: log(3x+7) + log(x-2) = log((3x+7)(x-2)) = 1 log(3x^2 + x - 14) = 1 What this means is that 10^1 = 3x^2 + x - 14, thus: 3x^2 + x - 14 = 10 3x^2 + x - 24 = 0 Factoring this is a little tricky, but you get: (x + 3)(3x - 8) = 0 Gonna flip the two around as we solve. 3x - 8 = 0 and x + 3 = 0 3x = 8 and x = -3 x = 8/3 and x = -3 There shouldn’t be any issues with x=8/3 if we check it: log(3(8/3)+7) + log((8/3)-2) = 1 log(8+7) + log(2/3) = 1 log(15) + log(2/3) = 1 log(15*2/3) = 1 log(5*2) = log(10) = 1 CORRECT So no issues there. However, we also have x=-3. The issue here is that plugging this into the original equation gives us logs of two negative numbers: log(3(-3)+7) + log((-3)-2) = log(-9+7) + log(-5) =log(-2) + log(-5) What to do here? Well, let’s just say the resulting complex number math gets messy. Plus, it means our answer will have an imaginary part, so it is not correct. The only valid answer is x = 8/3. Addendum: The reason why x=-3 comes out as a solution is because if you combine the logs above as follows: log(-2) + log(-5) = log(-2*-5) = log(10) = 1 Surprise win! No, sorry, but this doesn’t work when both numbers being combined are less than 0 (similar to dealing with negatives in radicals).
You CAN take the log of a negative number, but you will get a complex number, so x=-3 is an invalid solution. From what I can tell, plugging x=-3 into the original equation gives you something close to 1 + i*e, so you have a 1, but not by itself.
Answer: X = 8/3 or X = -3 ----------- log (3X+7)+ log (x-2)=1 Applying log a +log b= log ab, log(3X+7)+log(x-2)= log(3X+7)(x-2)=1 log 3x^2+x-14=1 3x^2+x-14=10 3x^2+x-24=0 (3x-8)(x+3)=0 Either 3x-8=0 OR x+3=0 If 3x-8=0, 3x=8 x= 8/3 If x+3=0, x=-3
log(3x+7)+log(x-2)=1
log{(3x+7)(x-2)}=1
10^log{(3x+7)(x-2)}=10^1
{(3x+7)(x-2)}=10
3x^2-6x+7x-14=10
3x^2+x-24
what are we solving for?
[-b+/-sqrt(b^2-4ac]/2a
a=3
b=1
c=-24
[-1+/-sqrt(1^2-4(3)(-24)]/(3(2))
[-1+/-sqrt(1+288)]/6
[-1+/-sqrt(289)]/6
[-1+/-17]/6
-18/6=-3
16/6
=8/3= 2.667
x= -3, 2.667
interim verify
3x^2+x-24=?0
3(-3)^2+(-3)-24=(0)
27-3-24=?0
27-27=❤0✔️
3(2.667)^2+(2.667)-24=?0
21.339+2.667-24=?0
24.006-24=~❤0✔️
sol.1 x=-3
sol.2 x=2.66
log(3x+7)+log(x-2)=1
log(3(-3)+7)+log(-3-2)=?1
log(-2)+log(-5)=?1
---undefined--
log(3x+7)+log(x-2)=1
log(3(2.66)+7)+log(2.66-2)=?1
log(7.98+7)+log(0.66)=?1
log(14.98)+log(0.66)=?1
1.17581 -0.1804=?1
0.9954~=❤1✔️
ANSWER x=2.666
log(3x + 7) + log(x - 2) = 1
Logarithm -> product property :
log_b(mn) = log_b(m) + log_b(n)
log(3x + 7)(x - 2) = 1
rewrite in exponential form :
10^1 = (3x + 7)(x - 2)
10 = 3x(x - 2) + 7(x - 2)
10 = 3x² - 6x + 7x - 14
10 -10 = 3x² - 6x + 7x - 14 - 10
0 = 3x² - 6x + 7x - 24
multiply the leading coefficient 3 with -24
0 = x² + x - 72
search for 2 numbers that if multiplied gives -72
and added together gives 1, the invisible 1 before the x
the numbers are -8 · 9 = -72 ∧ -8 + 9 = 1
(x - 8)(x + 9) = 0
divide the constant 8 and 9 by the leading coefficient 3
simplify the fractions if possible and multiply the denominators
with x
(x - 8/3)(x + 9/3) = 0
(3x - 8)(x + 3/1) = 0
(3x - 8)(x + 3) = 0
3x -8 = 0 ∨ x + 3 = 0
3x = 8
x = 8/3 ✅
x + 3 = 0
x = -3 --> logarithms of negative
numbers are undefined
the root {8/3} ✅
-------------------------------------------------------------------
check for extraneous solutions :
log(3x + 7) + log(x - 2) = 1
log(3(8/3) + 7) + log((8/3) - 2) = 1
log(8 + 7) + log(8/3 - 6/3) = 1
log(15) + log(2/3) = 1
log(15)(2/3) = 1
log(10) = 1 --> 10^1 = 10✅
---------------------------------------------------------
undefined because of the negative number
in the argument of the log
log(3x + 7) + log(x - 2)
log(3(-3) + 7) + log(-3 - 2)
log(-2) + log(-5) = 1
log(-2)(-5) = 1
log(10) = 1 --> 10^1 = 10
Oh, this is one of those times where you have to remember that identities work both ways. Here we go.
log(3x+7) + log(x-2) = 1
There’s a clever saying, “Everyone knows 1+1=2, but I KNOW 2=1+1.” Indeed, solving this hinges on understanding that:
log(m*n) = log(m) + log(n)
Also means:
log(m) + log(n) = log(m*n)
Here we have the sum of two logs with the same base, therefore we can combine them like so:
log(3x+7) + log(x-2) = log((3x+7)(x-2)) = 1
log(3x^2 + x - 14) = 1
What this means is that 10^1 = 3x^2 + x - 14, thus:
3x^2 + x - 14 = 10
3x^2 + x - 24 = 0
Factoring this is a little tricky, but you get:
(x + 3)(3x - 8) = 0
Gonna flip the two around as we solve.
3x - 8 = 0 and x + 3 = 0
3x = 8 and x = -3
x = 8/3 and x = -3
There shouldn’t be any issues with x=8/3 if we check it:
log(3(8/3)+7) + log((8/3)-2) = 1
log(8+7) + log(2/3) = 1
log(15) + log(2/3) = 1
log(15*2/3) = 1
log(5*2) = log(10) = 1 CORRECT
So no issues there. However, we also have x=-3. The issue here is that plugging this into the original equation gives us logs of two negative numbers:
log(3(-3)+7) + log((-3)-2) = log(-9+7) + log(-5)
=log(-2) + log(-5)
What to do here? Well, let’s just say the resulting complex number math gets messy. Plus, it means our answer will have an imaginary part, so it is not correct. The only valid answer is x = 8/3.
Addendum: The reason why x=-3 comes out as a solution is because if you combine the logs above as follows:
log(-2) + log(-5) = log(-2*-5) = log(10) = 1
Surprise win! No, sorry, but this doesn’t work when both numbers being combined are less than 0 (similar to dealing with negatives in radicals).
Complex solutions : ⅙±i√47/6
Negative 3 must be rejected because it makes at least one log term negative and logs of negative numbers are undefined. Therefore x equals 8/3 only.
You would think he would mention this in the video. At time index 2:18 he flashes "The answer is -3 and 8/3" on the screen; very misleading.
great lesson, thanks.
I agree that 8/3 is a valid solution
-3 works in the quadratic, but not in the original equation. So is it valid or not?
You CAN take the log of a negative number, but you will get a complex number, so x=-3 is an invalid solution. From what I can tell, plugging x=-3 into the original equation gives you something close to 1 + i*e, so you have a 1, but not by itself.
@@stevendebettencourt7651 Thanks. Read your other post, and kinda get the idea.
Answer:
X = 8/3 or X = -3
-----------
log (3X+7)+ log (x-2)=1
Applying
log a +log b= log ab,
log(3X+7)+log(x-2)=
log(3X+7)(x-2)=1
log 3x^2+x-14=1
3x^2+x-14=10
3x^2+x-24=0
(3x-8)(x+3)=0
Either 3x-8=0
OR
x+3=0
If 3x-8=0,
3x=8
x= 8/3
If x+3=0,
x=-3
(3x+7)(x-2)=10; 3x^2+x-24=0; x=(-1±17)/6=8/3 or -3
❤❤❤❤❤❤❤❤❤❤❤