Confused by wolfram alpha solution for solving log equation for y... Watch

mh1985
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A linear fit between "x" and "log_10(y)", and gets the following function:

log_10(y) = -1.4 x + 5.1

Write out the formula for the relationship between the original variables, x and y.

I tried solving this using wolfram alpha but can't understand their solution:

Their first step is:

x + 0.31021 log_10(y) = 3.64286

Now it looks like they've divided by 1.4 and moved the x to the LHS but 1/1.4 isn't 0.31021? it's 0.714285, I can't figure out where they got this from?

Thanks for the help
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Ateo
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(Original post by mh1985)
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Your interpretation of the output of wolfram alpha is incorrect. In wolfram alpha, log is used to denote log_e and not log_{10}.

\displaystyle \frac{1}{1.4 \cdot log_e(10)} is approximately 0.31021 .

You shouldn't need more than a calculator to do this question.

Note that log_a(b) = c \iff a^c = b.
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mh1985
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(Original post by Ateo)
Your interpretation of the output of wolfram alpha is incorrect. In wolfram alpha, log is used to denote log_e and not log_{10}.

\displaystyle \frac{1}{1.4 \cdot log_e(10)} is approximately 0.31021 .

You shouldn't need more than a calculator to do this question.

Note that log_a(b) = c \iff a^c = b.
Thanks for the answer
I changed it in wolfram alpha so it was using log_{10} not ln though?
I still don't understand why in the quote why we need to divide by the log ? \displaystyle \frac{1}{1.4 \cdot log_e(10)} ? All I can see to do is divide by 1.4?
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Ateo
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(Original post by mh1985)
Thanks for the answer
I changed it in wolfram alpha so it was using log_{10} not ln though?
I still don't understand why in the quote why we need to divide by the log ? \displaystyle \frac{1}{1.4 \cdot log_e(10)} ? All I can see to do is divide by 1.4?
You did input the base ten log indeed, but the output provided by wolfram was in base e. As for where what I wrote comes from, note that

\displaystyle log_{10}(x) = \frac{log_e(x)}{log_e(10)}

this is just the change of base rule for logarithms
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mh1985
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(Original post by Ateo)
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Ah thank you, that clears it up
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