Question 239135
{{{10^(2-5x)=793}}}
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:
{{{log((10^(2-5x)))=log((793))}}}
Now we can use the property of logarithms, {{{log(a, (p^q)) = q*log(a, (p))}}}, to move the exponent in the argument in front:
{{{(2-5x)log((10))=log((793))}}}
and since the log(10) = 1 (This is why I chose base 10 logarithms):
{{{2-5x = log((793))}}}
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:
{{{-5x = -2 + log((793))}}}
Divide both sides by -5:
{{{x = (-2 + log((793)))/(-5)}}}
or
{{{x = (2 - log((793)))/(5)}}}<br>
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:
{{{x = (2 - log((100*7.93)))/(5)}}}
Use the property of logarithms, {{{log(a, (p*q)) = log(a, (p)) + log(a, (q))}}}, to split the 100 and 7.93:
{{{x = (2 - (log((100)) + log((7.93))))/(5)}}}
Since {{{100 = 10^2}}}, {{{log((100)) = 2}}}:
{{{x = (2 - (2 + log((7.93))))/5}}}
By factoring out the 100, we now have a logarithm we can find in a table:
{{{x = (2 - (2 + 0.8993))/5}}}
I'll leave this to you to simplify.<br>
If you are not supposed to use a table, I do not see another way to get rid of the logarithm in
{{{x = (2 - log((793)))/5}}}
All we can do is manipulate the expression into possibly more desirable forms. As we saw before {{{2 = log((100))}}} so we substitute for the 2:
{{{x = (log((100)) - log((793)))/(5)}}}
Now we can use the property of logarithms, {{{log(a, (p)) - log(a, (q)) = log(a, (p/q))}}}, to combine the logarithms in the numerator:
{{{x = (log((100/793)))/(5)}}}
We can change the division by 5 into the multiplication by its reciprocal, 1/5:
{{{x = (1/5)(log((100/793)))}}}
Then we can use the previously used property involving exponents, {{{log(a, (p^q)) = q*log(a, (p))}}}, to move the number in front into the argument as an exponent:
{{{x = log(((100/793)^(1/5))))}}}
Writing the fractional exponent in radical form we get:
{{{x = log((root(5, (100/793))))}}}
<br>
Which is a "better" answer?
{{{x = (2 - log((793)))/(5)}}}
or
{{{x = log((root(5, (100/793))))}}}
I can't say for sure. I prefer the first one.