Question 718213
{{{log(10, (y))= -.75logx - 1.76}}}
To solve for y, all we have to do is rewrite the equation in exponential form. In general, {{{log(a, (p)) = q}}} is equivalent to {{{p = a^q}}}. In words, this pattern tells us that the argument of the log, p, is equal to the base of the log, a, with an exponent of the whole right-hand side of the logarithmic equation.<br>
Using this pattern, our log's argument, y, is equal to the base of our log, 10, with an exponent of the entire right-hand side:
{{{y = 10^(-.75logx - 1.76)}}}
which makes
y= 10^-.75logx-1.76 
almost right. You just need parentheses on the exponent:
y= 10^(-.75logx-1.76)