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Solve and find the domain of the equation:

 

\(\log_3 ((\log_{0.5}x)^2-(3\log_{0.5}x) +5)=2\)

michaelcai  Feb 12, 2018
 #1
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Solve for x over the real numbers:
log(3, log_0.5^2(x) - 3 log(0.5, x) + 5) = 2

log(3, log_0.5^2(x) - 3 log(0.5, x) + 5) = log(5 + 4.32809 log(x) + 2.08137 log^2(x))/log(3):
log(5 + 4.32809 log(x) + 2.08137 log^2(x))/log(3) = 2

Multiply both sides by log(3):
log(5 + 4.32809 log(x) + 2.08137 log^2(x)) = 2 log(3)

2 log(3) = log(3^2) = log(9):
log(5 + 4.32809 log(x) + 2.08137 log^2(x)) = log(9)

Cancel logarithms by taking exp of both sides:
5 + 4.32809 log(x) + 2.08137 log^2(x) = 9

Divide both sides by 2.08137:
2.40227 + 2.07944 log(x) + log^2(x) = 4.32408

Subtract 2.40227 from both sides:
2.07944 log(x) + log^2(x) = 1.92181

Add 1.08102 to both sides:
1.08102 + 2.07944 log(x) + log^2(x) = 3.00283

Write the left hand side as a square:
(log(x) + 1.03972)^2 = 3.00283
Take the square root of both sides:
log(x) + 1.03972 = 1.73287 or log(x) + 1.03972 = -1.73287

Subtract 1.03972 from both sides:
log(x) = 0.693147 or log(x) + 1.03972 = -1.73287

Cancel logarithms by taking exp of both sides:
x = 2. or log(x) + 1.03972 = -1.73287

Subtract 1.03972 from both sides:
x = 2. or log(x) = -2.77259

Cancel logarithms by taking exp of both sides:
x = 2                          or                     x = 0.0625

 

Domain:{x element R : x>0} (all positive real numbers) (assuming a function from reals to reals)

Guest Feb 13, 2018

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