Question 239135: 10^(2-5x)=793 please solve without a calculator
Found 2 solutions by JimboP1977, jsmallt9:
Answer by JimboP1977(311)   (Show Source):
Answer by jsmallt9(3758)  (Show Source):
You can put this solution on YOUR website! 
As far as I can tell, 793 is not an integral power of any integer. So the only way I know how to solve for x is to find the base 10 logarithm of each side:
Now we can use the property of logarithms,  , to move the exponent in the argument in front:
and since the log(10) = 1 (This is why I chose base 10 logarithms):
Now that x is no longer in an exponent we can use basic Algebra to solve for it. Add -2 to (or subtract 2 from) each side:
Divide both sides by -5:
or
If "without calculators" means you are allowed to use tables of logarithms instead (Do textbooks come with tables of logarithms in the back anymore?), then we could do the following:
Factor out 100 from 793:
Use the property of logarithms,  , to split the 100 and 7.93:
Since  ,  :
By factoring out the 100, we now have a logarithm we can find in a table:
I'll leave this to you to simplify.
If you are not supposed to use a table, I do not see another way to get rid of the logarithm in
All we can do is manipulate the expression into possibly more desirable forms. As we saw before  so we substitute for the 2:
Now we can use the property of logarithms,  , to combine the logarithms in the numerator:
We can change the division by 5 into the multiplication by its reciprocal, 1/5:
Then we can use the previously used property involving exponents, , to move the number in front into the argument as an exponent:
Writing the fractional exponent in radical form we get:
Which is a "better" answer?
or
I can't say for sure. I prefer the first one.
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