All SAT II Math I Resources
Example Questions
Example Question #61 : Sat Subject Test In Math I
Give the set of real solutions to the equation
(round to the nearest hundredth, if applicable)
The equation has no real solutions.
Using the Product of Powers Rule, then the Power of a Power Rule, rewrite the first term:
Substitute for ; the equation becomes
which is quadratic in terms of . The trinomial might be factorable using the method, where we split the middle term with integers whose product is and whose sum is . By trial and error, we find the integers to be and , so the equation can be rewritten as follows:
Factoring by grouping:
By the Zero Product Rule, one of these two factors must be equal to 0.
If , then .
Since , then substituting this as well as substituting back for , we get
,
and
If , then
Since , then substituting this as well as substituting back for , we get
, and
The solution set is therefore
Example Question #61 : Sat Subject Test In Math I
Give the set of real solutions to the equation
(round to the nearest hundredth, if applicable)
Using the Product of Powers Rule, then the Power of a Power Rule, rewrite the first term:
Substitute for ; the equation becomes
,
which is quadratic in terms of . The trinomial might be factorable using the method, where we split the middle term with integers whose product is and whose sum is 11. By trial and error, we find the integers to be 12 and , so the equation can be written as follows:
Factoring by grouping:
By the Zero Product Rule, one of these two factors must be equal to 0.
If , then .
Substituting back for , we get
.
This is impossible, since any power of a positive number must be positive.
If , then:
Substituting back for , we get
Since ,
it holds that , and , the only solution.
Example Question #61 : Sat Subject Test In Math I
Simplify
One of the properties of log is that
Applying that principle to this problem:
Simplifying the log base 10
Plug in the values to the first equation:
Example Question #21 : Mathematical Relationships
Evaluate:
An exponential base raised to the natural log will eliminate, leaving only the terms of the power. This is a log rule that can be used to simplify the expression.
Distribute the x variable through the binomial.
The answer is:
Example Question #15 : Exponents And Logarithms
By the Power of a Power and Product of Power Rules, we can rewrite this equation as
Substitute for ; the resulting equation is the quadratic equation
,
which can be written in standard form by subtracting from both sides:
The quadratic trinomial fits the perfect square trinomial pattern:
By the square root principle,
Substituting for :
Example Question #16 : Exponents And Logarithms
Solve for (round to the nearest hundredth):
Take the natural logarithm of both sides:
By Logarithm of a Power Rule, the above becomes
After distributing, solve for :
Factor out the left side, then divide:
Substituting the values of the logarithms:
This rounds to 0.45.
Example Question #12 : Exponents And Logarithms
Solve for :
No solution
, so the equation
can be rewritten as:
By the Power of a Power rule:
It follows that
Solving for :
Example Question #1 : Absolute Value
Define an operation on the set of real numbers as follows:
For any two real numbers
Evaluate the expression
Substitute in the expression:
Example Question #1 : Absolute Value
Simplify the following expression:
To simplify, we must first simplify the absolute values.
Now, combine like terms:
Example Question #2 : Absolute Value
Solve for .
To solve for x we need to make two separate equations. Since it has absolute value bars around it we know that the inside can equal either 7 or -7 before the asolute value is applied.
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