4^1-3b=8^2b+1

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sanman777
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4^1-3b=8^2b+1

Postby sanman777 » Mon Mar 23, 2009 4:27 pm

4^1-3b=8^2b+1

appreciate the help

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stapel_eliz
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Postby stapel_eliz » Mon Mar 23, 2009 5:05 pm

sanman777 wrote:4^1-3b=8^2b+1

appreciate the help

I would assume that the instructions were something along the lines of "solve the equation". Unfortunately, your formatting leaves something to be desired.... :oops:

Before we can know whether or not this can be solved algebraically, you'll need to provide clarification of what the equation is. For instance, does "8^2b+1" mean "82b + 1", "82b + 1", "82b + 1", or something else?

When you reply, please include a clear listing of your steps and reasoning so far. Thank you! :D

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Re: 4^(1-3b)=8^(2b+1)

Postby sanman777 » Mon Mar 23, 2009 5:16 pm

it says to solve algebraically

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stapel_eliz
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Postby stapel_eliz » Mon Mar 23, 2009 5:25 pm

sanman777 wrote:it says to solve algebraically

Okay; now we have the instructions. But what, exactly, is the equation?

When you reply, please show all your work so far. Thank you! :D

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Re: 4^1-3b=8^2b+1

Postby sanman777 » Mon Mar 23, 2009 5:39 pm

4^1-3b=8^2b+1

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stapel_eliz
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Postby stapel_eliz » Mon Mar 23, 2009 5:46 pm

sanman777 wrote:4^1-3b=8^2b+1

I'm going to guess that the above means the following:

. . . . .4^(1 - 3b) = 8^(2b + 1)

(If you're not wanting to use standard web-safe formatting, please use LaTeX, so people can understand your meaning. Thank you!)

The trick for solving this sort of exponential equation is to convert the bases to the same value. For instance:

. . . . .

Since 27 = 33 and 9 = 32, then:

. . . . .

. . . . .

The equation can then be restated as:

. . . . .

Equate the powers:

. . . . .

...and solve the resulting linear equation.

Your exercise works the exact same way. :wink:

sanman777
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Re: 4^1-3b=8^2b+1

Postby sanman777 » Mon Mar 23, 2009 5:52 pm

kk thanks alot


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