Log inequality question, help needed!!!

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Johnpeters
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0.8^(1-3x)>10 (greater than OR EQUAL TO not just greater than)

I used base of 10

log0.8 ^(1-3x)> log10
1-3x(log0.8)> log10
1-3x> (log10)/(log0.8)
-3x> (log10/log0.8)-1
x<((log10/log0.8)-1)/3
x< 3.77 (Less than OR EQUAL TO)

Could someone please tell me where I've gone wrong? I'm pretty sure the answer is x>3.77 rather than x<3.77. Was I not meant to flip the inequality when dividing by -3?
Thanks
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Enzo Ferrari
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By computing log(0.8), you will find out that it is negative. Hence, you also flip the inequality sign when dividing by log(0.8) to achieve your required result.
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TheBBQ
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(Original post by Johnpeters)
0.8^(1-3x)>10 (greater than OR EQUAL TO not just greater than)

I used base of 10

log0.8 ^(1-3x)> log10
1-3x(log0.8)> log10
1-3x> (log10)/(log0.8)
-3x> (log10/log0.8)-1
x<((log10/log0.8)-1)/3
x< 3.77 (Less than OR EQUAL TO)

Could someone please tell me where I've gone wrong? I'm pretty sure the answer is x>3.77 rather than x<3.77. Was I not meant to flip the inequality when dividing by -3?
Thanks
Log0.8 is a negative number , should have flipped the sign there too

Log(x), where x < 1 is negative, remember the log graph?
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Johnpeters
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Thank you, would never have thought of that!

And no I've never seen a log graph? We've done exponential graphs, is it anything to do with them?
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krisshP
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(Original post by Johnpeters)
Thank you, would never have thought of that!

And no I've never seen a log graph? We've done exponential graphs, is it anything to do with them?
Graph of lnx? Perhaps The BBQ was talking about this?
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davros
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(Original post by Johnpeters)
Thank you, would never have thought of that!

And no I've never seen a log graph? We've done exponential graphs, is it anything to do with them?

(Original post by krisshP)
Graph of lnx? Perhaps The BBQ was talking about this?
Every log function is the inverse of a corresponding exponential function for some base. If you've seen an exponential graph then you can get the inverse (log) graph by reflecting in y = x.

And whatever base you're using log 1 = 0 with log x < 0 when x < 1
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