# If Logx-6 base3=2y and Logx-7 base2=3y, show that x^2-13x+42=72y

## You have a typo.

the right side of your equation should say 72^y

log3(x-7) = 2y --> 3^(2y) = x-7
log2(x-6) = 3y --> 2^(3y) = x-6

then
LS = x^2 - 13x + 42
= (x-6)(x-7)
= ( 2^(3y) ) ( 3^(2y) )
= (8^y)(9^y)
= 72^y
= RS

## Well, let's have some fun with math!

First, let's start by rewriting the given equations:

Logx-6 base3 = 2y ----> (1)
Logx-7 base2 = 3y ----> (2)

Now, let's express the logarithmic equations in exponential form:

3^(2y) = x - 6 ----> (3)
2^(3y) = x - 7 ----> (4)

To eliminate the variables in equations (3) and (4), let's try something silly. We're going to square equation (3) and cube equation (4):

(3^(2y))^2 = (x - 6)^2
(2^(3y))^3 = (x - 7)^3

Simplifying, we get:

9^y = (x - 6)^2 ----> (5)
8^y = (x - 7)^3 ----> (6)

Now, let's get rid of the exponent by taking the logarithm of both equations:

log(9^y) = log((x - 6)^2)
log(8^y) = log((x - 7)^3)

Using the logarithmic properties, we can rewrite these equations as:

y * log(9) = 2 * log(x - 6)
y * log(8) = 3 * log(x - 7)

Dividing equation (5) by 2 and equation (6) by 3:

log(x - 6) = (y * log(9)) / 2 ----> (7)
log(x - 7) = (y * log(8)) / 3 ----> (8)

Now, let's do something clownish and add equations (7) and (8):

log(x - 6) + log(x - 7) = (y * log(9)) / 2 + (y * log(8)) / 3

Using logarithmic properties, we can combine the logarithms:

log((x - 6)(x - 7)) = (3y * log(9) + 2y * log(8)) / 6

log((x - 6)(x - 7)) = (9y * log(9) + 12y * log(8)) / 18

Now, let's raise both sides to the power of 10:

(x - 6)(x - 7) = 10^((9y * log(9) + 12y * log(8)) / 18)

Now multiply both sides by 18:

18(x - 6)(x - 7) = 10^(9y * log(9) + 12y * log(8))

Finally, we end up with:

18(x - 6)(x - 7) = (9^y)(8^y)

Expanding and simplifying:

18(x^2 - 13x + 42) = 72y

Dividing both sides by 18:

x^2 - 13x + 42 = 4y

Wait a second...I made an error! We didn't get 72y exactly. So, it seems my clownish calculations have backfired this time. My apologies, this humor-filled journey led us astray!

In conclusion, I was unable to prove that x^2 - 13x + 42 equals 72y. I better stick to making people laugh rather than complex mathematical equations.

## To prove that x^2 - 13x + 42 = 72y given:

1. Logₓ₋₆ₕₑ₃ = 2y
2. Logₓ₋₇₂ = 3y

We can start by expressing the logarithmic equations in exponential form.

1. For the equation Logₓ₋₆ₕₑ₃ = 2y:
This can be rewritten as ₃^(2y) = x₋₆. (Exponential form of logarithm)

2. For the equation Logₓ₋₇₂ = 3y:
This can be rewritten as ₂^(3y) = x₋₇. (Exponential form of logarithm)

Now, we can substitute x₋₆ and x₋₇ into the quadratic equation x² - 13x + 42 to prove that it equals 72y.

Substituting x₋₆ in the equation:
₃^(2y) = x₋₆
Squaring both sides:
₃^(4y) = x₋₆²

Substituting x₋₇ in the equation:
₂^(3y) = x₋₇
Cubing both sides:
₂^(9y) = x₋₇³

Now, let's substitute these values into the quadratic equation:

x² - 13x + 42 = 72y

Replace x₋₆ with ₃^(4y) and x₋₇ with ₂^(9y):

(₃^(4y))² - 13(₃^(4y)) + 42 = 72y

Simplifying, we get:

₉^(4y) - 13(₃^(4y)) + 42 = 72y

After simplifying further, we obtain:

81y² - 117y + 42 = 72y

Rearranging the terms:

81y² - 189y + 42 = 0

Now, we can solve this quadratic equation for y using factoring, completing the square, or the quadratic formula.

Once we find the values of y, we can substitute them back into the original logarithmic equations to find the corresponding values of x. Then, we can substitute the values of x and y in the quadratic equation x² - 13x + 42 = 72y to verify that it holds true.

(Note: The actual equation obtained in step 10 may vary, depending on any simplification or substitution made during the process).

## To prove that x^2 - 13x + 42 = 72y using the given logarithmic equations, we need to find the value of x that satisfies both equations and then substitute that value into the equation x^2 - 13x + 42.

Let's start by solving the first logarithmic equation: Logx-6 base3 = 2y.

We can rewrite the equation in exponential form: 3^(2y) = x - 6.

Now, let's solve the second logarithmic equation: Logx-7 base2 = 3y.

Rewriting in exponential form: 2^(3y) = x - 7.

Now, we have two equations:

1) 3^(2y) = x - 6
2) 2^(3y) = x - 7

To find the value of x that satisfies both equations, we can set the right-hand sides of the equations equal to each other:

x - 6 = x - 7

Simplifying, we get: 1 = 0

Since 1 does not equal 0, these two equations cannot hold simultaneously. Therefore, there is no value of x that satisfies both equations as given.

Thus, we cannot prove that x^2 - 13x + 42 = 72y using the given logarithmic equations.