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

Complex numbers, rational functions, logarithms, sequences and series, matrix operations, etc.
sanman777
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### 4^1-3b=8^2b+1

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

appreciate the help

stapel_eliz
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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....

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!

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

it says to solve algebraically

stapel_eliz
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sanman777 wrote:it says to solve algebraically

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

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

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

stapel_eliz
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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:

. . . . .$27^{2x}\, =\, 9^{x\, -\, 4}$

Since 27 = 33 and 9 = 32, then:

. . . . .$27^{2x}\, =\, (3^3)^{2x}\, =\, 3^{3\times 2x}\, =\, 3^{6x}$

. . . . .$9^{x\, -\, 4}\, =\, (3^2)^{x\, -\, 4}\, =\, 3^{2\times(x\, -\, 4)}\, =\, 3^{2x\, -\, 8}$

The equation can then be restated as:

. . . . .$3^{6x}\, =\, 3^{2x\, -\, 8}$

Equate the powers:

. . . . .$6x\, =\, 2x\, -\, 8$

...and solve the resulting linear equation.

Your exercise works the exact same way.

sanman777
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kk thanks alot