SOLUTION: solve for x: log base 4(x^2+3x)=1+log base 4(x=5)

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Question 118887: solve for x:
log base 4(x^2+3x)=1+log base 4(x=5)

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
log%284%2C%28x%5E2%2B3x%29%29=1%2Blog%284%2C%28x%2B5%29%29

Step 1: Add -log%284%2C%28x%2B5%29%29 to both sides

log%284%2C%28x%5E2%2B3x%29%29-log%284%2C%28x%2B5%29%29=1

Step 2: Apply log%28a%2CC%29-log%28a%2CD%29=log%28a%2C%28C%2FD%29%29

log%284%2C%28%28x%5E2%2B3x%29%2F%28x%2B5%29%29%29=1

Step 3: Apply the y=log%28b%2Cx%29 => b%5Ey=x equivalency

4%5E1=%28%28x%5E2%2B3x%29%2F%28x%2B5%29%29

Step 4: Multiply both sides by (x + 5)

4x%2B20=x%5E2%2B3x

Step 5: Collect terms and put in standard form

x%5E2-x-20=0

Step 6: Factor

%28x-5%29%28x%2B4%29=0

x=5 or x=-4

Check:
log%284%2C%28%285%29%5E2%2B3%285%29%29%29=1+%2B+log%284%2C%285%2B5%29%29
log%284%2C40%29=1+%2B+log%284%2C10%29 but since 4%5E1=4, 1=log%284%2C4%29 so
log%284%2C40%29=log%284%2C4%29%2Blog%284%2C10%29
log%284%2C40%29=log%284%2C%284%2A10%29%29, so x=5 checks

log%284%2C%28%28-4%29%5E2%2B3%28-4%29%29%29=1+%2B+log%284%2C%28-4%2B5%29%29
log%284%2C4%29=1+%2B+log%284%2C1%29 but since 4%5E1=4, 1=log%284%2C4%29 so
log%284%2C4%29=log%284%2C4%29%2Blog%284%2C1%29
log%284%2C4%29=log%284%2C%284%2A1%29%29, so x=-4 checks