Question 348510
An equation of this form:
log(expression) = 0ther-expression
is usually solved by rewriting it in exponential form. In general, {{{log(a, (p)) = q}}} is equivalent to {{{a^q = p}}}. In your case:
{{{log(n, (39870)) = 28}}}
is equivalent to:
{{{n^28 = 39870}}}<br>
To solve this equation we want to end up with an equation of the form:
n = expression
The "n" above has no visible exponent. Nevertheless the "n" does have an exponent. Any "invisible" exponent is always understood to be a 1. So, even though we don't usually write it this way, what we are looking for is:
{{{n^1}}} = experession.<br>
So how to we change the {{{n^28}}} that we have into the {{{n^1}}} that we want? Thinking through our rules for exponents and thinking about when we get a 1 as an answer, we may be able to realize that<ul><li>the rule for exponents, {{{(a^x)^y = a^(x*y)}}} gives us a situation where we multiply exponents, and</li><li>Multiplying reciprocals <i>always</i> results in 1!</li></ul>
So in order to change the {{{n^28}}} to {{{n^1}}} we will need to raise it to the reciprocal of 28, 1/28, power. And if we raise the left side to the 1/28 power, we must also raise the right side to the same power:
{{{(n^28)^(1/28) = (39870)^(1/28)}}}
which simplifies to
{{{n = 39870^(1/28)}}}
which, in radical form, is
{{{n = root(28, 39870)}}}
This is an exact expression for the solution. If you need a decimal approximation, then we can use out calclators. If your calculator has buttons for parentheses, "(" and ")", then you can just type:
39870^(1/28)
If your calculator does not have buttons for parentheses, then first calculate 1/28 (approximately 0.0357142857) and use this decimal for the exponent:
39870^0.0357142857
Either way you should get appoximately 1.46.