Solving x + lg x = a, where "lg" is "log, base 2" (ans is "8" for a=11, but how...?)

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Solving x + lg x = a, where "lg" is "log, base 2" (ans is "8" for a=11, but how...?)

If lg is log base 2, how do I solve the following:


x + lg x = a


If a is 11 then the answer is 8 (8 + lg 8 = 8 + 3 = 11) but how do I solve it algebraically?
 
… lg is log base 2 …

x + lg x = a

… how do I solve it algebraically?
This equation cannot be solved for x algebraically.

x is an algebraic function, and log2(x) is a transcendental function. When we create new functions by combining algebraic and transcendental functions, we often need higher levels of math to work with them.

After calculus, some students learn about the LambertW function. This function was created (designed) to allow solutions for equations like the one you posted.

Therefore, the solution for x looks like this (written in terms of the LambertW function, the constants e and a, and the natural logarithm of 2):

\(\displaystyle x = \dfrac{\text{LambertW}[ln(2) \cdot e^{a \cdot ln(2)}]}{ln(2)}\)

This case of defining new math in order to deal with an expression that cannot be handled otherwise, is similar to what mathematicians did when dealing with square roots of negative numbers. They created (designed) a new number called the imaginary unit: \(\displaystyle i\). If you've learned that \(\displaystyle i\) is defined as the square root of -1, then you probably know also that we need \(\displaystyle i\) to be able to solve those quadratic equations for which the Discriminant is negative. :cool:
 
That's very interesting. It does indeed solve the equation (I used WolframAlpha to get the value of LambertW).


It's interesting that such a simple-looking equation would take a special effort to solve. It makes me wonder why the exponential function "transcends" algebra.


I assume the ln(2)'s are there to convert from base e to base 2, so if the original question was about base e they would all become 1.


Thank you so much for your answer.
 
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