for the second equation I did, (2^4x) x 2 = 3^x (16^x) x 2 = 3^x (3^x)/(16^x) = 2 (3/16)^x = 2 Taking log base (3/16) on both sides x = log base 3/16 (2) x = -0.414 (which is the same as yours) great problems, looking forward to more!
Im supposed to solve using only natural log. that has been my problem finding examples of people solving with that and not log! Stuff like 2^(5x+4)=3^(3x-2) can you just write ln instead of log?
Since we have the change-of-base rule, it is arbitrary whether you write ln(8)/ln(2) or log(8)/log(2). Both produce the same result. You can solve any problem involving logs, using either natural log or log base ten. Or even some completely different base like log base 2. To do the problem you provided using natural log: Given: 2^(5*x+4) = 3^(3*x - 2) Take the natural log of both sides: ln(2^(5*x + 4)) = ln(3^(3*x - 2)) Use the log property, ln(a^b) = b*ln(a) to pull the exponents out in front: (5*x + 4)*ln(2) = (3*x - 2)*ln(3) Expand, move constants to the right, and variables to the left: 5*x*ln(2) + 4*ln(2) = 3*x*ln(3) - 2*ln(3) 5*x*ln(2) - 3*x*ln(3) = -2*ln(3) - 4*ln(2) Factor the left: [5*ln(2) - 3*ln(3)]*x = -2*ln(3) - 4*ln(2) Isolate x: x=[-2*ln(3) - 4*ln(2)]/[5*ln(2) - 3*ln(3)] This can simplify to: x = -ln(144)/ln(32/27), which evaluates to about -29.25
for the second equation I did,
(2^4x) x 2 = 3^x
(16^x) x 2 = 3^x
(3^x)/(16^x) = 2
(3/16)^x = 2
Taking log base (3/16) on both sides
x = log base 3/16 (2)
x = -0.414 (which is the same as yours)
great problems, looking forward to more!
I like how this was explained so clearly and simply. Crisp, without needless complications.
He teaches better in one second than many teachers in 1 hour.
2:00 his humor is crazy
"Surprising, right?"
Hahahaha
so that's when "log that's not base e" is important 😮
It's cool how the answer to the second one still has 1,2,3,4 in it.
I am highly impressed sir
I love this guy. It's just funny because if he would have just took the ln of 2 and 3 in the first place he wouldn't have to re-write the answer
i just used the property of logarithms, where you put the exponent in front of the logarithm
(3x+1)log2 = xlog4
3xlog2 + log2 = xlog4
3xlog2 - xlog4 = -log2
x(3log2-log4) = -log2
x = -log2/3log2-log4
It also the x is ²log (3/16)
Well explained
That is a good one!!!
crazy guy
Im supposed to solve using only natural log. that has been my problem finding examples of people solving with that and not log! Stuff like 2^(5x+4)=3^(3x-2) can you just write ln instead of log?
Since we have the change-of-base rule, it is arbitrary whether you write ln(8)/ln(2) or log(8)/log(2). Both produce the same result. You can solve any problem involving logs, using either natural log or log base ten. Or even some completely different base like log base 2.
To do the problem you provided using natural log:
Given: 2^(5*x+4) = 3^(3*x - 2)
Take the natural log of both sides:
ln(2^(5*x + 4)) = ln(3^(3*x - 2))
Use the log property, ln(a^b) = b*ln(a) to pull the exponents out in front:
(5*x + 4)*ln(2) = (3*x - 2)*ln(3)
Expand, move constants to the right, and variables to the left:
5*x*ln(2) + 4*ln(2) = 3*x*ln(3) - 2*ln(3)
5*x*ln(2) - 3*x*ln(3) = -2*ln(3) - 4*ln(2)
Factor the left:
[5*ln(2) - 3*ln(3)]*x = -2*ln(3) - 4*ln(2)
Isolate x:
x=[-2*ln(3) - 4*ln(2)]/[5*ln(2) - 3*ln(3)]
This can simplify to:
x = -ln(144)/ln(32/27), which evaluates to about -29.25
What is 2 log base2 of 3
Those nikes bruh
Isn’t (2^2)^x equal 4^x?
yes
Nice Shoes and nice video also
Beautiful
wow