Solving Exponentials using Logs: 27^(x-1) = 81^(2x+3)

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53andfab
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Solving Exponentials using Logs: 27^(x-1) = 81^(2x+3)

Postby 53andfab » Sat Nov 13, 2010 6:30 pm

Solve:

27^(x-1) = 81^(2x+3)

Thanks

nicknation
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Joined: Wed Mar 11, 2009 3:28 pm

Re: Solving Exponentials using Logs: 27^(x-1) = 81^(2x+3)

Postby nicknation » Sat Nov 13, 2010 11:06 pm

53andfab wrote:Solve: 27^(x-1) = 81^(2x+3)

Change 27 and 81 to powers of 3. Then set the powers equal.

They show how to do that here. You don't need logs.


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