Solving y = 1/4 log(2) (8x – 56)^16 – 12

Question

Hello I would like to know if whether I simplified y = 1/4 log2 (8x – 56)^16 – 12 correctly or not. Here are my steps:

1. 12 = log2 (8x – 56)^4 => I moved the 1/4 to the power using the log laws. 1/4 x 16 equals 4. I also moved the -12 to the other side which equals +12.

2. Log2(8x)^4 – log2(56)^4 = 12 => Since (8x-56) has the same log base I rewrote it as two terms.

3. Log2(8x/56)^4 = 12 => Using the log laws I divided the 8x with 56.

4. Log2(x/7)^4 = 12 => I simplified the fraction.

5. Log2(x^4/7^4 ) = 12 => I applied the power of 4 to the fraction.

6. 2^12 = (x^4/7^4 ) => I converted the logarithm to an exponential equation

7. 4096/2401 = x^4 => I carried the 2401 (7^4) to the other side. Dividing the 2^12 with 7^4.

8. ∜(4096/2401) = x => I fourth rooted both sides.

9. x = 8/7 = I arrived with this fraction.

My question is, did I solve the equation correctly? Thank you!

Answer 1

You had a good start. $12 = \log_2 (8x – 56)^4$ can be rewritten as $12 = 4 \log_2 |8x – 56|$ or $\log_2 |8x – 56|=3$ which leads to $|8x-56|=8$. Can you finish?

$y+12 = \log_2 (8x – 56)^4$ can be simplified to $2^{y/4}=|x-7|$.

Answer 2

The separation you did in step 2 is not valid. Log does not distribute over subtraction. Also (though your answer is way off by then), in step 7 you divided when you meant to multiply.

You can always check your work by plugging your answer back into the original equation you were trying to solve.