+619 Solve and find the domain of the equations:
\(x^2 \cdot \log_x 27 \cdot \log_9 x=x+4\)
and
\(2\log_{4x}x^3=5 \log _{2x}x\)
Sorry for so many logarithm problems!
I'm just really bad at them
Solve for x:
(log(27) x^2)/log(9) = x + 4
Subtract x + 4 from both sides:
(log(27) x^2)/log(9) - x - 4 = 0
Divide both sides by log(27)/log(9):
-(log(9) x)/log(27) + x^2 - (4 log(9))/log(27) = 0
Add (4 log(9))/log(27) to both sides:
x^2 - (log(9) x)/log(27) = (4 log(9))/log(27)
Add (log^2(9))/(4 log^2(27)) to both sides:
-(log(9) x)/log(27) + x^2 + (log^2(9))/(4 log^2(27)) = (log^2(9))/(4 log^2(27)) + (4 log(9))/log(27)
Write the left hand side as a square:
(x - log(9)/(2 log(27)))^2 = (log^2(9))/(4 log^2(27)) + (4 log(9))/log(27)
Take the square root of both sides:
x - log(9)/(2 log(27)) = sqrt((log^2(9))/(4 log^2(27)) + (4 log(9))/log(27)) or x - log(9)/(2 log(27)) = -sqrt((log^2(9))/(4 log^2(27)) + (4 log(9))/log(27))
Add log(9)/(2 log(27)) to both sides:
x = sqrt((log^2(9))/(4 log^2(27)) + (4 log(9))/log(27)) + log(9)/(2 log(27)) or x - log(9)/(2 log(27)) = -sqrt((log^2(9))/(4 log^2(27)) + (4 log(9))/log(27))
Add log(9)/(2 log(27)) to both sides:
| x = sqrt((log^2(9))/(4 log^2(27)) + (4 log(9))/log(27)) + log(9)/(2 log(27)) or x = log(9)/(2 log(27)) - sqrt((log^2(9))/(4 log^2(27)) + (4 log(9))/log(27))
x = -4/3 and x = 2
Domain: {x element R : 0 1} (assuming a function from reals to reals)
Solve for x over the real numbers:
(2 log(x^3))/log(4 x) = (5 log(x))/log(2 x)
Cross multiply:
2 log(2 x) log(x^3) = 5 log(x) log(4 x)
Subtract 5 log(x) log(4 x) from both sides:
2 log(2 x) log(x^3) - 5 log(x) log(4 x) = 0
Transform 2 log(2 x) log(x^3) - 5 log(x) log(4 x) into a polynomial with respect to log(x):
log^2(x) - 4 log(2) log(x) = 0
Factor log(x) and constant terms from the left hand side:
-log(x) (4 log(2) - log(x)) = 0
Multiply both sides by -1:
log(x) (4 log(2) - log(x)) = 0
Split into two equations:
4 log(2) - log(x) = 0 or log(x) = 0
Subtract 4 log(2) from both sides:
-log(x) = -4 log(2) or log(x) = 0
Multiply both sides by -1:
log(x) = 4 log(2) or log(x) = 0
4 log(2) = log(2^4) = log(16):
log(x) = log(16) or log(x) = 0
Cancel logarithms by taking exp of both sides:
x = 16 or log(x) = 0
Cancel logarithms by taking exp of both sides:
x = 16 or x = 1
Domain: {x element R : 0 1} (assuming a function from reals to reals)
+129852 First one
x^2 * logx 27 * log9 x = x + 4 we can write
x^2 (log 27 / log x ) ( log x / log 9) = x + 4
x^2 (log 27 / log 9) = x + 4
x^2 ( log 3^3 / log3^2) = x + 4
x^2 *(3/2) = x + 4
(3/2)x^2 - x - 4 = 0
3x^2 - 2x - 8 = 0 factor
(3x + 4) ( x - 2) = 0
Setting each factor to 0 and solve for x and we have that x = -4/3 or x = 2
Only the second solution is good.....the first would mean that we were taking the log of a negative number in the original equation
