5
To solve for ( x ) in logarithmic equations, you can use the property that if ( \log_b(a) = c ), then ( a = b^c ). First, isolate the logarithmic expression, and then rewrite the equation in exponential form. Finally, solve for ( x ) by performing any necessary algebraic operations. For example, if you have ( \log_2(x) = 3 ), you can rewrite it as ( x = 2^3 ), which simplifies to ( x = 8 ).
To solve 10.005 - 18, you can rewrite the operation as 10.005 + (-18). Performing the subtraction, you get 10.005 - 18 = -7.995. Therefore, the answer is -7.995.
To subtract integers in algebra, you can use the rule of adding the opposite. This means that when you subtract an integer, you add its opposite. For example, to solve ( a - b ), you can rewrite it as ( a + (-b) ). This approach helps simplify the operation and makes it easier to find the result.
associative
b²
A = 5 + 3*b To solve for b, rewrite as b = (A - 5) / 3
c=a-b
l= A/ W
a+b=
To solve 10.005 - 18, you can rewrite the operation as 10.005 + (-18). Performing the subtraction, you get 10.005 - 18 = -7.995. Therefore, the answer is -7.995.
To subtract integers in algebra, you can use the rule of adding the opposite. This means that when you subtract an integer, you add its opposite. For example, to solve ( a - b ), you can rewrite it as ( a + (-b) ). This approach helps simplify the operation and makes it easier to find the result.
b = y - mx.
associative
b y = xlog(b) + log(y) = log(x)
pa/b = (pa)1/b = bth root of (pa)
b²
If: a = b+c+d Then: c = a-b-d