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How can I solve an exponent in a equation using base 10 logarithm tables? For an example, $$a = b^x$$ can be written as $$ \log_{10} a= x\log_{10}b $$ $$ x = \frac{ \log_{10}b }{\log_{10}a} $$ After this point I can refer values for $\log a$ and $\log b$ from the table. From this point how can I solve to get $x$. Can I subtract values since log division is subtraction??? Or should I take antilog. I'm Kinda stuck here Can Anyone help?

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    $\begingroup$ If you have the values for $\log a$ and $\log b$, you just need to divide the value for $x$ ... $\endgroup$ – Michael Burr Jan 3 '18 at 15:27
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    $\begingroup$ If you have value table use it in this stage, you have $\log a/\log b$ and not $\log(a/b)$ so you can't use antilog nor make it subtraction $\endgroup$ – Holo Jan 3 '18 at 15:28
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    $\begingroup$ Be careful, log division is $\log(a/b)$ and that can be rewritten as subtraction. However, division of logs is $\log(a)/\log(b)$ and is not subtraction. $\endgroup$ – Michael Burr Jan 3 '18 at 15:28
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    $\begingroup$ The identity that is applicable is $\frac{\log_{10}b}{\log_{10}a}=\log_ab$. However, depending on the table you have, you are probably better off keeping it this way, or even splitting the logs further as $\log_{10}b=\frac{\ln b}{\ln 10}$ and $\log_{10}a=\frac{\ln a}{\ln 10}$, if the table is of natural logarithms. $\endgroup$ – user517969 Jan 3 '18 at 15:30
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It seems like you're pretty much done. For example, let's say you've found $\log_{10}b = 6$ and $\log_{10}a = 2$. Then $x = \frac{6}{2} = 3$.

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  • $\begingroup$ thnxx.. But What if I get bar values How to divide then?? $\endgroup$ – saha isu Jan 3 '18 at 15:30
  • $\begingroup$ @sahaisu I'm not sure what you mean. Can you give me an example of a bar value? $\endgroup$ – BallBoy Jan 3 '18 at 15:32
  • $\begingroup$ If we assume the same values for a and b log a*10^-3 = 3bar.2 $\endgroup$ – saha isu Jan 3 '18 at 15:34
  • $\begingroup$ @sahaisu My example works if $a=10^2$. Then $a \cdot 10^{-3} = 10^{-1}$, so $\log_{10}(a\cdot 10^{-3}) = -1$. What does "3bar.2" mean? $\endgroup$ – BallBoy Jan 3 '18 at 15:42
  • $\begingroup$ Ahh yeah. But If we take a value for a such as 0.02 then referring log tables value for two is 0.3010 And for 0.02 is $$ \bar 2.3010 $$ So when solving using bar values such as above How can I divide? $\endgroup$ – saha isu Jan 3 '18 at 15:50

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