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# Last logarithm problems

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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

Feb 12, 2018

### 3+0 Answers

#1
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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)

Feb 13, 2018
#2
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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)

Feb 13, 2018
#3
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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   Feb 13, 2018