Please help. I have not had algebra for thirty years. How do you solve the following equation?

2^x=2x

Thanks for the help!

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- Sep 19th 2009, 07:06 PMRussWExponential Equation
Please help. I have not had algebra for thirty years. How do you solve the following equation?

2^x=2x

Thanks for the help! - Sep 19th 2009, 07:21 PMenjam
2^x = 2x

ln(2^x) = ln(2x)

xln(2) = ln(2) + ln (x)

x = [ln (2) + ln (x)] / ln (2)

x = 1 + ln(x)/ln(2)

From here, by simple inspection, we can see that x can be either 1 or 2.

when x = 1, we have 1 = 1 + ln(1)/ln(2) = 1 + 0

when x = 2, we have 2 = 1 + ln(2)/ln(2) = 1 + 1 - Sep 20th 2009, 05:49 AMRussW
Thanks for your response. Other than by inspection, is there a way to solve a more generic equation. For example, solve for x when a^x=bx.

- Sep 20th 2009, 05:50 AMe^(i*pi)
There are iterative methods but none analytical ones