solve in logs to base 10

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June 3rd 2012, 04:24 AM
sluggerbroth

solve in logs to base 10

(2^x) - 6/(2^-x)=6
June 3rd 2012, 04:40 AM
Prove It

Re: solve in logs to base 10

Quote:

Originally Posted by sluggerbroth

(2^x) - 6/(2^-x)=6

$\displaystyle \begin{align*} 2^x - \frac{6}{2^{-x}} &= 6 \\ 2^x - 6\cdot 2^x &= 6 \\ -5\cdot 2^x &= 6 \\ 2^x &= -\frac{6}{5} \end{align*}$

This does not have any solutions in the real numbers, because $\displaystyle \begin{align*} 2^x > 0 \end{align*}$ for all $\displaystyle \begin{align*} x \end{align*}$ .
June 3rd 2012, 04:44 AM
Plato

Re: solve in logs to base 10

Quote:

Originally Posted by sluggerbroth

(2^x) - 6/(2^-x)=6

If the problem is $2^x-\frac{6}{2^{-x}}=6$ then there is no solution.

If that is not the question, then what is it?
June 3rd 2012, 07:17 AM
Soroban

Re: solve in logs to base 10

Hello, sluggerbroth!

Quote:

(2^x) - 6/(2^-x) = 6

I suspect that the problem is: . $2^x - \frac{6}{2^x} \:=\:6$

Multiply by $2^x:\;(2^x)^2 - 6 \:=\:6\cdot2^x$

And we have: . $(2^x)^2 - 6\cdot2^x - 6 \:=\:0$ . . . a quadratic in $2^x$

Quadratic Formula: . $2^x \:=\:\frac{6\pm\sqrt{60}}{2} \:=\:3\pm\sqrt{15}$

Take logs, base 10: . $\log(2^x) \:=\:\log(3\pm\sqrt{15})$

. . . . . . . . . . . . . . $x\cdot\log(2) \:=\:\log(3\pm\sqrt{15})$

n . . . . . . . . . . . . . . . . . . $\boxed{\;x \:=\:\frac{\log(3 + \sqrt{15})}{\log(2)}\;}$

Note that $3 - \sqrt{15}$ is negative.

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