find x if
a) 1/81=3^2x-1
b)log2x+ 3/log2x = 4
since 81 = 3^4, we have
3^-4 = 3^(2x-1)
-4 = 2x-1
x = -3/2
logx + 3/logx = 4
log^2(x) - 4logx + 3 = 0
(logx-1)(logx-3) = 0
logx = 1,3
x = 2,8
check:
1+3/1 = 4 yes
3+3/3 = 4 yes
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a) To find x in the equation 1/81 = 3^(2x-1), we will use logarithms.
First, we can rewrite the equation using the standard logarithmic form for exponentiation:
81 = 3^(2x - 1)
To eliminate the base 3 exponent, we take the common logarithm (log base 10) of both sides:
log(81) = log(3^(2x - 1))
Using the logarithmic property that log(b^c) = c * log(b), we can simplify the equation:
log(81) = (2x - 1) * log(3)
Next, we can calculate the logarithms:
log(81) ≈ 1.908
log(3) ≈ 0.477
Now, we can solve for x by isolating it:
2x - 1 = log(81) / log(3)
Dividing both sides by 2 and adding 1 to both sides:
2x = (log(81) / log(3)) + 1
Finally, divide both sides by 2 to solve for x:
x = [(log(81) / log(3)) + 1] / 2
By substituting the respective logarithmic values, we can find the value of x.
b) To find x in the equation log(2x) + 3 / log(2x) = 4, we will use properties of logarithms.
First, let's simplify the equation by multiplying both sides by log(2x):
log(2x) * [log(2x) + 3 / log(2x)] = 4 * log(2x)
Simplifying the left side using the logarithmic property log(a) + log(b) = log(a * b):
log(2x) * [log(2x) + 3 - log(2x)] = 4 * log(2x)
The term log(2x) on the left side cancels out, so we're left with:
log(2x) * 3 = 4 * log(2x)
Now, we can cancel out log(2x) by dividing both sides of the equation:
3 = 4
Since 3 does not equal 4, the equation has no solution.