SAT II Math II : Exponents and Logarithms

Study concepts, example questions & explanations for SAT II Math II

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Example Questions

Example Question #41 : Sat Subject Test In Math Ii

Give the set of real solutions to the equation

(round to the nearest hundredth, if applicable)

Possible Answers:

The equation has no solution.

Correct answer:

Explanation:

Using the Product of Powers Rule, then the Power of a Power Rule, rewrite the first two terms strategically:

Substitute  for ; the equation becomes

Factor this as 

 

by finding two integers whose product is 3 and whose sum is . Through some trial and error we find , so we can write

By the Zero Product Rule, one of these two factors must be equal to 0.

If , then .

Substituting  back for :

.

If , then .

Substituting  back for :

Both can be confirmed to be solutions by substitution.

Example Question #41 : Sat Subject Test In Math Ii

Give the solution set:

Possible Answers:

The equation has no solution.

Correct answer:

Explanation:

Rewrite by taking advantage of the Product of Powers Property and the Power of a Power Property:

Substitute  for ; the resulting equation is the quadratic equation

which can be written in standard form by subtracting  from both sides:

The trinomial can be factored by the  method, Look for two integers with sum  and product ; by trial and error, we find they are , so the equation can be rewritten and solved by grouping:

By the Zero Product Property, one of these factors must be equal to 0. 

Either

Substituting  back for :

Or:

Substituting  back for :

The solution set, as can be confirmed by substituting in the equation, is .

Example Question #11 : Exponents And Logarithms

Solve for :

Possible Answers:

Correct answer:

Explanation:

 and , so, 

can be rewritten as

Applying the Power of a Power Rule,

Example Question #22 : Mathematical Relationships

Solve the equation:  

Possible Answers:

Correct answer:

Explanation:

Rewrite the base of the right side.

Simplify the right side.

Add 6 on both sides.

Divide by 6 on both sides.

The answer is:  

Example Question #23 : Mathematical Relationships

Solve:  

Possible Answers:

Correct answer:

Explanation:

Change the base of the left side to base two.

The equation becomes:

Set the exponents equal since they have similar bases.

Divide by 2 on both sides.

The answer is:  

Example Question #51 : Sat Subject Test In Math Ii

Simplify:

Possible Answers:

Correct answer:

Explanation:

Start by distributing the exponent in both the numerator and the denominator. Recall that when an exponent is raised to the exponent, you will need to multiply the two numbers together.

Next, recall that when you have numbers with different exponents, but the same base, subtract the exponent found in the denominator from the exponent in the numerator.

Recall that you can flip the fraction to make the exponents positive.

Example Question #52 : Sat Subject Test In Math Ii

Solve .

Possible Answers:

Correct answer:

Explanation:

The first thing we need to do is find a common base.  This can be tricky to do, but guessing and checking a little shows that:

Plugging that back in to the original equation:

Now that our bases are the same, we can cancel them:

From here, it's much easier to solve using simple algebra:

Example Question #53 : Sat Subject Test In Math Ii

Solve .

Possible Answers:

Correct answer:

Explanation:

First, we gather all the constants on one side of the equation:

Next, we rewrite the equation in exponential form:

Now we can simplify the exponent:

And finally, divide:

Example Question #54 : Sat Subject Test In Math Ii

Solve .

Possible Answers:

Correct answer:

Explanation:

Start by combining log terms.  Remember, if you subtract logs, you divide the terms inside them:

Now we can rewrite the equation in exponential form:

Finally, we need to get the variable in the numerator, and then alone:

Example Question #55 : Sat Subject Test In Math Ii

Solve .

Possible Answers:

Correct answer:

Explanation:

First, we combine log terms by subtracting them.  Remember, when you subtract logs, you divide the terms inside them:

Now, because the bases of the logs match on either side of the equation, we can cancel them out:

From here, we use simple algebra to solve:

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