Logs and Exponentials?

Question

I’m unsure on how I would get two possible answers for this?? Please help

Answer 1

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1 + 2 log₃(x) = log₃(28x - 9)

We can move the log on the left side to the right by subtracting that term from both sides:

1 = log₃(28x - 9) - 2 log₃(x)

Next, I'll move the "2" that's outside the second log inside by putting it as an exponent to what's inside:

1 = log₃(28x - 9) - log₃(x²)

The difference of two logs with the same base is the same as the log of the quotient, so:

1 = log₃[(28x - 9) / x²]

Now if we make both sides of the equation an exponent over a base of 3 this will cancel out the log and leave only what's inside the log:

3¹ = (28x - 9) / x²

3 = (28x - 9) / x²

Multiply both sides by x² then we can simplify this as a quadratic:

3x² = 28x - 9

3x² - 28x + 9 = 0

I'll use the quadratic equation:

x = [ -b ± √(b² - 4ac)] / (2a)

x = [ -(-28) ± √((-28)² - 4(3)(9))] / (2 * 3)

x = [ 28 ± √(784 - 108)] / 6

x = [ 28 ± √(676)] / 6

x = (28 ± 26) / 6

x = 2/6 and 54/6

x = 1/3 and 9

Testing the solutions to make sure they are correct:

1 + 2 log₃(x) = log₃(28x - 9)

1 + 2 log₃(1/3) = log₃(28 * 1/3 - 9) and 1 + 2 log₃(9) = log₃(28 * 9 - 9)

1 + 2 log₃(1/3) = log₃(28/3 - 9) and 1 + 2 log₃(9) = log₃(252 - 9)

1 + 2 log₃(1/3) = log₃(28/3 - 27/3) and 1 + 2 log₃(9) = log₃(243)

1 + 2 log₃(1/3) = log₃(1/3) and 1 + 2 log₃(9) = log₃(243)

1/3, 9, and 243 are all powers of 3:

1 + 2 log₃(3⁻¹) = log₃(3⁻¹) and 1 + 2 log₃(3²) = log₃(3⁵)

1 + 2(-1) = -1 and 1 + 2(2) = 5

1 - 2 = -1 and 1 + 4 = 5

-1 = -1 and 5 = 5

TRUE and TRUE

Both are solutions, so your answer is D from your multiple choice.

Answer 2

1 + 2log₃x = log₃(28x – 9)

log₃3 = 1

log₃3 + 2log₃x = log₃(28x – 9)

log₃3 + log₃x² = log₃(28x – 9)

log₃(3x²) = log₃(28x – 9)

3x² = 28x – 9

3x² – 28x + 9 = 0

quadratic equation:

to solve ax² + bx + c = 0

x = [–b ± √(b²–4ac)] / 2a

x = [28 ± √(28²–4•3•9)] / 6

x = [28 ± √(784–108)] / 6

x = [28 ± √(676)] / 6

x = [28 ± 26] / 6

x = 1/3, 9

Answer 3

1  =log_b3(3) 

log_b3(3) + 2*log_b3(x)   = log_b3(28x-9)

log_b3(3)  + log_b3(x^2)  =log_b3(28x-9) 

log_b3(  3 *x^2)     =  log_b3( 28x-9)   

therefore 

3x^2  = 28x -9

3x^2 -28x + 9 = 0  

factor or use quadratic formula 

(3x -a)(x - b)     

(3x - 1) (x -9)  =0     

so either 

3x -1 = 0 

or 

x -9 = 0 

3x = 1  

x= 1/3 

or 

x = 9 

so D is correct