Condensing logarithms I got to this: \(\displaystyle{2}{{\log}_{{{10}}}{x}^{{{\frac{{{1}}}{{{2}}}}}}}+{3}{{\log}_{{{10}}}{x}^{{{\frac{{{1}}}{{{3}}}}}}}\) Now, usually

Harley Ayers

Harley Ayers

Answered question

2022-03-25

Condensing logarithms
I got to this: 2log10x12+3log10x13
Now, usually you bring the exponent the the front and that would yield:
12(2)log10x+13(3)log10x=log10x+log10x=2log10x
And that's it?

Answer & Explanation

disolutoxz61

disolutoxz61

Beginner2022-03-26Added 12 answers

12(2)log10x+13(3)log10x2log10x+3log10x
122=1 so do 133.
So the correct thing is
12(2)log10x+13(3)log10x=log10x+log10x=2log10x
Bring the exponent inside the log if you like.
The other way around: bring 1/2 into log10x to become log10x2=??.

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