Question 713038
I'm assuming that the expression is:
{{{log(7, (10))}}}
If I'm wrong then you'll have to re-post your question making the expression more understandable.<br>
Did you post the problem exactly as it was given to you? I ask because it is impossible to find the exact value/number that equals {{{log(7, (10))}}}. If it was possible to find the exact number for logarithms like this we would never have to learn how to work with this syntax. Why would we ever use such cumbersome syntax as {{{log(a, (p))}}} if we could replace them with numbers?!<br>
The best we can do for something <i>exactly</i> equal to {{{log(7, (10))}}} is to rewrite it in terms of logarithms of other bases. We can use the base conversion formula, {{{log(a, (p)) = log(b, (p))/log(b, (a))}}}, to change this base 7 log into an expression involving any other base.<br>
For two reasons I am going to use the change of base formula to change the base 7 log into an expression of base 10 logs:<ul><li>The 10 in the base 7 log</li><li>Our calculators "know" base 10 logs so we will be able to use our result to find the decimal approximation.</li></ul>Using the base conversion formula to convert
{{{log(7, (10))}}}
into an expression of base 10 logs we get:
{{{log((10))/log((7))}}}
By definition, any log whose base matches its argument is equal to 1. So log(10) = 1 and our fraction simplifies to:
{{{1/log((7))}}}
This is another exact expression (but not a "value") for {{{log(7, (10))}}}.<br>
We can use this expression, and our calculators, to find a decimal approximation:
{{{1/0.84509804001425683071221625859264}}}
which simplifies to:
1.1832946624549383268179285616469<br>
Note 1: No matter how many decimal places we use, we will never get {{{log(7, (10))}}} exactly right. Even 1.1832946624549383268179285616469 is just a very, very close, rounded-off approximation of {{{log(7, (10))}}}.
Note 2: Since they're not exact anyway, feel free to round off these long decimals.