Question 421232
If the given point is on the graph of the given equation, then the coordinates of the point must be a solution to the equation. (IOW: The coordinates must fit the equation.) So:
{{{(3) = log(2, (5(2) + b))}}}
We can solve this for b, First let's simplify:
{{{3 = log(2, (10 + b))}}}
Now we rewrite the equation in exponential form. In general {{{log(a, (p)) = q}}} is equivalent to {{{a^q = p}}}. Using this pattern on our equation we get:
{{{2^3 = 10+b}}}
which simplifies to:
{{{8 = 10 + b}}}
Now we just subtract 10:
{{{-2 = b}}}