SOLUTION: I need help on how to solve a logarithmic Equation. The question is 3 Log (base 2) x+1 = 7 When I tried putting this equation into exponential form, I got 2^7 = (x+1)^3

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Question 242094: I need help on how to solve a logarithmic Equation. The question is
3 Log (base 2) x+1 = 7
When I tried putting this equation into exponential form, I got

2^7 = (x+1)^3. However, when I solved for this, I did not end up with
the answer that is needed. I ended up looking up the answer in the back of the book so that i could see if i could backtrack from there, but it was simply 4.
Thanks,
dmc93x@yahoo.com
Found 2 solutions by jim_thompson5910, solver91311:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
3%2Alog%282%2C%28x%29%29%2B1=7 Start with the given equation.


3%2Alog%282%2C%28x%29%29=7-1 Subtract 1 from both sides.


3%2Alog%282%2C%28x%29%29=6 Combine like terms.


log%282%2C%28x%29%29=6%2F3 Divide both sides by 3.


log%282%2C%28x%29%29=2 Reduce


x=2%5E2 Convert to exponential form.


x=4 Square 2 to get 4.


So the solution is x=4

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


Well, the back of the book missed it by that much.

If , then , and . But . That tells us that has to be a little bit bigger than 4, i.e. precisely the cube root of 128 minus 1.







Now, if perchance your text book gave you instructions to round to the nearest whole number, then 4 is, in fact, the correct answer. You might also want to re-check the answer in the back to make sure the decimal part of the answer didn't wrap to the next line. Barring any of that, write letters to the publisher and the author suggesting that they hire a proofreader next time they get it into their head to publish a textbook.

John