Remember
![log(a^z)=zlog(a)](https://tex.z-dn.net/?f=log%28a%5Ez%29%3Dzlog%28a%29)
so
notice that 200=2*2*2*5*5, so x cannot be a whole number
take the log base 10 of both sides
![log_{10}(2^x)=log{10}(200)](https://tex.z-dn.net/?f=log_%7B10%7D%282%5Ex%29%3Dlog%7B10%7D%28200%29)
![xlog_{10}(2)=log{10}(200)](https://tex.z-dn.net/?f=xlog_%7B10%7D%282%29%3Dlog%7B10%7D%28200%29)
divide both sides by
![log_{10}(2)](https://tex.z-dn.net/?f=log_%7B10%7D%282%29)
![x= \frac{log_{10}(200)}{log_{10}(2)}](https://tex.z-dn.net/?f=x%3D%20%5Cfrac%7Blog_%7B10%7D%28200%29%7D%7Blog_%7B10%7D%282%29%7D%20)
if we used
![log_{2}](https://tex.z-dn.net/?f=log_%7B2%7D)
instead of
![log_{10}](https://tex.z-dn.net/?f=log_%7B10%7D)
we would get
![x= log_{2}(200)](https://tex.z-dn.net/?f=x%3D%20log_%7B2%7D%28200%29%20)
because
Answer:
x= 7/3
Step-by-step explanation:
y=2x+5
5x-2=2x+5
transfer 2x to the left side and -2 to the right side you get
5x-2x=2+5
as it transfers the sign changes
3x=7
x=7/3
Let’s change the equation into
y=mx+b
5x-6y=13 Subtract 5x on both sides
-6y=-5x+13 Then divide by -6 on both sides
Y=5/6x-13/6
m will be your slope
so:
Slope=5/6
Answer:
6
Step-by-step explanation:
First observe that if
,
![(a + b)^2 = a^2 + 2ab + b^2 \\\\ \implies a + b = \sqrt{a^2 + 2ab + b^2} = \sqrt{a^2 + ab + b(a + b)} \\\\ \implies a + b = \sqrt{a^2 + ab + b \sqrt{a^2 + ab + b(a+b)}} \\\\ \implies a + b = \sqrt{a^2 + ab + b \sqrt{a^2 + ab + b \sqrt{a^2 + ab + b(a+b)}}} \\\\ \implies a + b = \sqrt{a^2 + ab + b \sqrt{a^2 + ab + b \sqrt{a^2 + ab + b \sqrt{\cdots}}}}](https://tex.z-dn.net/?f=%28a%20%2B%20b%29%5E2%20%3D%20a%5E2%20%2B%202ab%20%2B%20b%5E2%20%5C%5C%5C%5C%20%5Cimplies%20a%20%2B%20b%20%3D%20%5Csqrt%7Ba%5E2%20%2B%202ab%20%2B%20b%5E2%7D%20%3D%20%5Csqrt%7Ba%5E2%20%2B%20ab%20%2B%20b%28a%20%2B%20b%29%7D%20%5C%5C%5C%5C%20%5Cimplies%20a%20%2B%20b%20%3D%20%5Csqrt%7Ba%5E2%20%2B%20ab%20%2B%20b%20%5Csqrt%7Ba%5E2%20%2B%20ab%20%2B%20b%28a%2Bb%29%7D%7D%20%5C%5C%5C%5C%20%5Cimplies%20a%20%2B%20b%20%3D%20%5Csqrt%7Ba%5E2%20%2B%20ab%20%2B%20b%20%5Csqrt%7Ba%5E2%20%2B%20ab%20%2B%20b%20%5Csqrt%7Ba%5E2%20%2B%20ab%20%2B%20b%28a%2Bb%29%7D%7D%7D%20%5C%5C%5C%5C%20%5Cimplies%20a%20%2B%20b%20%3D%20%5Csqrt%7Ba%5E2%20%2B%20ab%20%2B%20b%20%5Csqrt%7Ba%5E2%20%2B%20ab%20%2B%20b%20%5Csqrt%7Ba%5E2%20%2B%20ab%20%2B%20b%20%5Csqrt%7B%5Ccdots%7D%7D%7D%7D)
Let
and
. It follows that
![a+b = x = \sqrt{x \sqrt{x \sqrt{x \sqrt{\cdots}}}}](https://tex.z-dn.net/?f=a%2Bb%20%3D%20x%20%3D%20%5Csqrt%7Bx%20%5Csqrt%7Bx%20%5Csqrt%7Bx%20%5Csqrt%7B%5Ccdots%7D%7D%7D%7D)
Now let
, so
. Solving for
,
![a^2 + a - 4x = 0 \implies a = \dfrac{-1 + \sqrt{1+16x}}2](https://tex.z-dn.net/?f=a%5E2%20%2B%20a%20-%204x%20%3D%200%20%5Cimplies%20a%20%3D%20%5Cdfrac%7B-1%20%2B%20%5Csqrt%7B1%2B16x%7D%7D2)
which means
![a+b = \dfrac{1 + \sqrt{1+16x}}2 = \sqrt{4x + \sqrt{4x + \sqrt{4x + \sqrt{\cdots}}}}](https://tex.z-dn.net/?f=a%2Bb%20%3D%20%5Cdfrac%7B1%20%2B%20%5Csqrt%7B1%2B16x%7D%7D2%20%3D%20%5Csqrt%7B4x%20%2B%20%5Csqrt%7B4x%20%2B%20%5Csqrt%7B4x%20%2B%20%5Csqrt%7B%5Ccdots%7D%7D%7D%7D)
Now solve for
.
![x = \dfrac{1 + \sqrt{1 + 16x}}2 \\\\ 2x = 1 + \sqrt{1 + 16x} \\\\ 2x - 1 = \sqrt{1 + 16x} \\\\ (2x-1)^2 = \left(\sqrt{1 + 16x}\right)^2](https://tex.z-dn.net/?f=x%20%3D%20%5Cdfrac%7B1%20%2B%20%5Csqrt%7B1%20%2B%2016x%7D%7D2%20%5C%5C%5C%5C%202x%20%3D%201%20%2B%20%5Csqrt%7B1%20%2B%2016x%7D%20%5C%5C%5C%5C%202x%20-%201%20%3D%20%5Csqrt%7B1%20%2B%2016x%7D%20%5C%5C%5C%5C%20%282x-1%29%5E2%20%3D%20%5Cleft%28%5Csqrt%7B1%20%2B%2016x%7D%5Cright%29%5E2)
(note that we assume
)
![4x^2 - 4x + 1 = 1 + 16x \\\\ 4x^2 - 20x = 0 \\\\ 4x (x - 5) = 0 \\\\ 4x = 0 \text{ or } x - 5 = 0 \\\\ \implies x = 0 \text{ or } \boxed{x = 5}](https://tex.z-dn.net/?f=4x%5E2%20-%204x%20%2B%201%20%3D%201%20%2B%2016x%20%5C%5C%5C%5C%204x%5E2%20-%2020x%20%3D%200%20%5C%5C%5C%5C%204x%20%28x%20-%205%29%20%3D%200%20%5C%5C%5C%5C%204x%20%3D%200%20%5Ctext%7B%20or%20%7D%20x%20-%205%20%3D%200%20%5C%5C%5C%5C%20%5Cimplies%20x%20%3D%200%20%5Ctext%7B%20or%20%7D%20%5Cboxed%7Bx%20%3D%205%7D)
(we omit
since
is not true)