(1/6)to the power of 3x = 36 to the power of 2-x
(1/6)^(3x) = 36 ^(2-x)
hint: 1/6 = 6^(-1) & 36 = 6^2
6^[-1(3x)] =6^[2(2-x)] :- -3x =4-2x x=4
To solve the equation (1/6)^(3x) = 36^(2-x), we need to first simplify both sides of the equation.
Let's start by simplifying the exponents separately.
On the left side of the equation, we have (1/6)^(3x). To simplify this, we can rewrite it using the rule of exponents:
(1/6)^(3x) = (1^(3x))/(6^(3x)) = 1/(6^(3x))
On the right side of the equation, we have 36^(2-x). We can rewrite this using the rule of exponents as well:
36^(2-x) = (6^2)^(2-x) = 6^(2(2-x)) = 6^(4-2x)
Now, our equation becomes:
1/(6^(3x)) = 6^(4-2x)
To simplify further, we can get rid of the denominator by multiplying both sides of the equation by 6^(3x):
(1/(6^(3x))) * (6^(3x)) = (6^(4-2x)) * (6^(3x))
1 = 6^(4-2x) * 6^(3x)
Next, we use the rule of exponents that states when multiplying two numbers with the same base, you can add their exponents:
1 = 6^(4-2x+3x)
Simplifying the exponent:
1 = 6^(4+x)
Now, we need to solve for x. To do this, we need to take the logarithm of both sides of the equation. Let's use the natural logarithm, denoted as ln:
ln(1) = ln(6^(4+x))
Using the property of logarithms that states ln(a^b) = b * ln(a):
0 = (4 + x) * ln(6)
Now, we can solve for x by isolating it:
4 + x = 0
x = -4
So, the solution to the equation (1/6)^(3x) = 36^(2-x) is x = -4.