Solve for x:
Log base 3 square root of x +1/2 = 3log base x 3
I find your lack of parentheses ... disturbing
I shall solve
log_3(√x) + 1/2 = 3/log_3(x)
Now, assuming base 3 for the logs,
1/2 logx + 1/2 = 3/logx
log^2(x) + logx - 6 = 0
(logx+3)(logx-2) = 0
logx = -3 or 2
x = 1/27 or 9
you can easily check that both solutions work
To solve for x in the given equation:
Log base 3 of the square root of x + 1/2 = 3log base x of 3
we can simplify it step by step:
Step 1: Rewrite the square root as a fractional exponent:
Log base 3 of (x^(1/2)) + 1/2 = 3log base x of 3
Step 2: Apply the power rule of logarithms:
1/2 * log base 3 of x + 1/2 = 3 * (log base 3 of 3) / (log base 3 of x)
Since log base 3 of 3 is equal to 1, the equation becomes:
1/2 * log base 3 of x + 1/2 = 3 / (log base 3 of x)
Step 3: Multiply both sides by 2 to eliminate the fraction:
log base 3 of x + 1 = 6 / (log base 3 of x)
Step 4: Multiply both sides by (log base 3 of x) to get rid of the denominator:
(log base 3 of x)^2 + (log base 3 of x) = 6
Step 5: Substitute y = log base 3 of x:
y^2 + y - 6 = 0
Step 6: Factor the quadratic equation:
(y + 3)(y - 2) = 0
Step 7: Solve for y:
y + 3 = 0 or y - 2 = 0
y = -3 or y = 2
Step 8: Substitute y back into the equation y = log base 3 of x and solve for x:
For y = -3:
log base 3 of x = -3
x = 3^(-3)
x = 1/27
For y = 2:
log base 3 of x = 2
x = 3^2
x = 9
Therefore, the solutions for x are x = 1/27 and x = 9.