What is absolute value? can you solve problems using the absolute value?

Mathematics

2 answers:

timofeeve [1]2 years ago

4 0

Answer:

Absolute value is a numbers distance from zero. It keeps positive numbers and turns negative numbers positive.

Yes, you can solve problems using the absolute value. For example,

|4| - |1| = 4-1 = 3

vesna_86 [32]2 years ago

3 0

An absolute value is positive value of any value. So the abs value of -28 is 28. The abs value of 67 is 67. Makes sense?

If it were |27-3| for example, treat the inside of a abs as parenthesis, so you must complete PEMDAS inside of it to reduce the equation to |24|, unless you wanted it to become |27| - |3|.

For functions, this becomes slightly different and more difficult, especially when adding a variable such as x. Look below for a sample equation.

|2x-3|=1

This equation will actually have (and most others) 2 solutions for x. To find these, you’ll need to multiply the inside of the abs by -1 for one equation, and leave it as it is for the other!

2x-3=1 -(2x-3)=1

Now you have to solve BOTH equations to get your correct x-value answers.

For the first listed equation:

2x=4

x=2

For the second listed equation:

-2x+3=1

-2x=-2

x=-1

So you get the x-values -1 and 2 which both make the parent function true!

You might be interested in

If logb2=x and logb3=y, evaluate the following in terms of x and y:

Alja [10]

$log_b{162} = x + 4y\\\\log_b324 = 2x+4y\\\\log_b\frac{8}{9} = 3x-2y\\\\\frac{log_b27}{log_b16} = 3y-4x$

Solution:

Given that,

$log_b2 = x\\\\log_b3 = y --------(i)$

Use the following log rules

Rule 1: $log_b(ac) = log_ba + log_bc$

Rule 2: $log_b\frac{a}{c} = log_ba - log_bc$

Rule 3: $log_ba^c = clog_ba$

$a) log_b{162}$

Break 162 down to primes:

$162 = 2^1 \times 3^4$

$log_b{162} =log_b 2^1. 3^4\\\\By\ rule\ 1\\\\ log_b{162} = log_b 2^1 +log_b 3^4\\\\By\ rule\ 3\\\\1log_b2 + 4log_b3\\\\1x+4y\\\\x+4y$

Thus we get,

$log_b162 = x + 4y$

$b) log_b 324$

Break 324 down to primes:

$324 = 2^2 \times 3^4$

$log_b324 = log_b 2^2.3^4\\\\By\ rule\ 1\\\\log_b324 = log_b2^2 + log_b3^4\\\\By\ rule\ 3\\\\log_b324 = 2log_b2 + 4log_b3\\\\From\ (i)\\\\log_b324 = 2x + 4y$

$c) log_b\frac{8}{9}$

By rule 2

$log_b\frac{8}{9} = log_b8 - log_b9\\\\log_b\frac{8}{9} = log_b 2^3 - log_b3^2\\\\By\ rule\ 3\\\\log_b\frac{8}{9} = 3 log_b2 - 2log_b3\\\\From\ (i)\\\\log_b\frac{8}{9} = 3x - 2y$

$d) \frac{log_b27}{log_b16}$

By rule 2

$\frac{log_b27}{log_b16} = log_b27 - log_b16\\\\ \frac{log_b27}{log_b16} = log_b3^3 - log_b2^4\\\\By\ rule\ 2\\\\ \frac{log_b27}{log_b16} = 3log_b3 - 4log_b2 \\\\From\ (i)\\\\\frac{log_b27}{log_b16} = 3y - 4x$