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Logarithms Question

This is the equation 7(2^n − 1). It asks to show that n > (log 200 007 − log 7) / log 2. I'm just not sure how to get to log 200007 because when i work it out it's just 200000. I'm sure its something to do with manipulating the 1 but not sure how to do it. My current working is:

log7(2n^-1)>log200000
log7 + log 2^n - log1 > log200000
(log1=0) so log7+nlog2>log200000
nlog2>log200000 - log7
n>(log200000 - log7)/ log 2
Original post by AQASUX
This is the equation 7(2^n − 1). It asks to show that n > (log 200 007 − log 7) / log 2. I'm just not sure how to get to log 200007 because when i work it out it's just 200000. I'm sure its something to do with manipulating the 1 but not sure how to do it. My current working is:

log7(2n^-1)>log200000
log7 + log 2^n - log1 > log200000
(log1=0) so log7+nlog2>log200000
nlog2>log200000 - log7
n>(log200000 - log7)/ log 2

Hi AQASUX

Starting with 7(2^n - 1) > 200000:

1) divide by 7
2^n - 1 > 200000/7

2) Add 1
2^n > 200000/7 +1
2^n > 200007/7

3) Take logs of both sides
n log 2 > log(200007/7)
n log 2 > log 200007 - log 7

4) Divide by log 2
n > (log 200007 - log 7) / log 2

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