Question #e10e6

1 Answer
Jul 10, 2016

Answer:

#m=-100#

Explanation:

The only way to get the answer

#m = -100#

is to have

#log(-m) + 2 = 4#

as your starting equation. As you know, the common log, which is denoted #log#, is actually the log base #10#. This means that your starting equation can be rewritten as

#log_10 (-m) + 2 = 4#

The first thing to do here is isolate the log on one side of the equation by adding #-2# to both sides

#log_10 (-m) + color(red)(cancel(color(black)(2))) - color(red)(cancel(color(black)(2)))= 4 -2#

#log_10(-m) = 2#

The log function is actually undefined for negative numbers when working with real numbers. This tells you that #-m >= 0#, which implies that #m <= 0#.

Now, the log function is the inverse operation to exponentiation. This means that you're looking for a number that is equal to the base, which in your case is #color(red)(10)#, raised to the power of the result, which is #color(blue)(2)#.

#log_color(red)(10)(color(blue)(-m)) = color(darkgreen)(2)#

Can thus be rewritten as

#color(red)(10)^color(darkgreen)(2) = color(blue)(-m)#

Since #10^2 = 100#, you will have

#100 = -m implies m = color(green)(|bar(ul(color(white)(a/a)color(black)(-100)color(white)(a/a)|)))#

As predicted, #m<=0#.