Solving the following equation: \(\displaystyle{2}^{{{x}}}+{{\log}_{{{10}}}{x}}-{2}={0}\)

Milton Robertson

Milton Robertson

Answered question

2022-03-31

Solving the following equation:
2x+log10x2=0

Answer & Explanation

Regan Gallegos

Regan Gallegos

Beginner2022-04-01Added 9 answers

Define f(x)=2x+logx2 and note f is increasing on the interval (0,) (since both 2x and logx) are increasing. Then there can be at most one root - it's easy to see that f(1)=0, so x=1 is the unique solution.
Carter Lin

Carter Lin

Beginner2022-04-02Added 13 answers

Try a few values and see what you get for this part:
2x+log10x2
For example, for x=1, you'll get 0 as 2x+log10x=2+0=2
Now, if x>1, then the 2x+log10x will be bigger than 2 so those can be removed. For 0<x<1, look at how the log now is negative and thus the exponent on the first term has to be greater than one. At least that how I'd look at it.

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