Solve:

2^(5x-6) = 7

My work:
log^(5x-6) = log7
5x - 6(log2) = log7
5x = log7 + 6(log2)
x = (log7 + log2^6) / 5

And textbook answer:
(log7) / (log2)

What did I do wrong?

2 answers

you are wrong and the textbook answer, the way you stated is, is wrong.

2^(5x-6) = 7
log[2^(5x-6)] = log7
(5x-6)log2 = log7
5x-6 = log7/log2
5x = log7/log2 + 6
x = (log7/log2 + 6)/5
Without using a calculator, find the value of the following logarithmic expression.

log7 1
Similar Questions
  1. Can someone see if I got these correct?1. Solve log5 x = 3. x = 125 2. log3 (x - 4) > 2. x > 13 3. Evaluate log7 49. log7 49 = 2
    1. answers icon 2 answers
    1. answers icon 2 answers
  2. Given log2=0.3010, log3=0.4771, log5=0.6690, and log7=0.8451...Find a.) log8 b) log5/7 c.) log1.5 d.) log3/14 e) log12 I didn't
    1. answers icon 2 answers
  3. Estimate the value of the logarithm to the nearest tenth.log7 (55) Options --------- A. 0.5 B. 2.1 C. -2.6 6.0
    1. answers icon 2 answers
more similar questions