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The domain is the set of possible input values. Usually "x" is the input variable so the domain is sometimes the set of possible values for x.

When finding domains you want to make sure certain "no-no's" cannot happen. These "non-no's" include (but are not limited to):

• Zero denominators,
• Negative radicands (expressions within a radical are called radicands) of even-numbered roots (square roots, 4th roots, 6th roots, etc.).
• Zero or negative arguments to logarithms.

In expression (a), you have both an even-numbered root, the square root, and a logarithm. So you must ensure that the radicand of the square root, "x", can never be negative: . And you must ensure that the argument of the logarithm, 10-x, can never be zero or negative (IOW it must be positive): . Putting these two together we have:
and
Solving the second one for x we get:
and
(Important tip: Always read inequalities like thees from where the variable is. That means we read the first one from left to right and the second one from right to left!) These inequalities tell use that x must be greater than or equal to 0 and less than (but not equal to) 10. This is the domain of your expression.

For expression (b) I'm not sure if you really mean "eye-en" which means nothing to me or "el-en" which means natural logarithm. If you really meant "In" then I don't know how to find the domain since I don't know what "In" means or whether it has some limits on its arguments.

If you mean "el-en", ln, then please be more careful in your posts. Tutors are less likely to help when problems are unclear.

For ln(x)+ln(2-x) we have two logarithms. we must ensure that both arguemtns remain positive:
and
Solving the second one we get:
and
This tells us that x must be greater than (but not equal to) 0 and x must be less than (but not equal to) 2. This is the domain of expression (b) if they were ln's and not In's.

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Please repost your problem. Use parentheses around arguments to logarithms and use additional parentheses as needed to make your expressions/equations unambiguous.

Here's some of the ways your expression could be interpreted:




etc.

Tutors are more likely to answer your questions if they are clearly stated.

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Once you understand what logarithms are this problem becomes very easy.

The idea behind logarithms is that it is possible to take any positive number (other than 1) and, if you raise it to the correct power, get any other positive number. For example, it is possible to raise 12 to some power and get 1/2. Or raise 3 to some power and get 153. The tricky part is finding out wht these exponents are. What power of 12 results in 1/2? And what power of 3 results in 153?

Logarithms are used to express these exponents. For example represents the exponent for 12 that results in 1/2. And represents the exponent for 3 that results in 153.

Now let's look at

Let's look at just the exponent: . This is "the exponent for 4 that results in 5" or "the power of 4 that results in 5". And where do we find ? Answer: As an exponent for 4! Since your expression raises 4 to the power of 4 that results in 5, your expression is simply a 5!

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Solving equations like this, where the variable is in the argument of a logarithm, usually starts by transforming the equation into one of the following forms:
log(expression) = other-expression
or
log(expression) = log(other-expression)

With the "non-log" term of 5/6 on the right side the second form, which is all logarithms, will be more difficult to achieve. So we will aim for the first form. The first form requires s single logarithm and your equation has two. Somehow we need to find a way to combine the two into one.

If the two logarithms were like terms then we could add them. But terms with logarithms are like terms only if both the bases and the arguments are the same. For example: (Just like z + 3z = 4z!) Your two logarithms have difference bases, 9 and 3, and different arguments, and x. So we cannot add them,

Another way to combine two logarithms into one is to use a property of logarithms. There are two such properties:

• 
• 

These properties require that the bases be the same and that the coefficients of the logs are 1's. Your two logarithms have bases that are different. However bases of logarithms can be changed using the base conversion formula: . We can use this formula to convert the base 9 log into a base 3 log:

And since , giving us:

We have made progress. The bases of the logarithms are equal. But we cannot use the property to combine the logs until we get rid of that 2 somehow. Since dividing by 2 is the same as multiplying by 1/2 we can rewrite the equation as:

This helps because there is yet a third property of logarithms, , which allows us to move the coefficient of a logarithm into the argument as an exponent:

The rule for exponents when raising a power to a power is to multiply the exponents. And 2*(1/2) is 1!

We can now combine the logarithms. We can use the property or we can even add them now since the two terms are now like terms (the bases and arguments are now the same!). Adding them we get:

Dividing by 2 we get:

We have finally reached the first form! The next step with this form is to rewrite the equation in exponential form. In general is equivalent to . Using this pattern on your equation we get:

This is an exact expression for the solution to your equation. You may recall that fractional exponents can also be written in radical form:

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First let's isolate the base, (x+1), and its exponent, 4/3, on one side of the equation by dividing both sides by 3:

Next we want to find a way to change the exponent from 4/3 to a 1. We can do so by combining all of the following ideas:

• One is allowed to raise both sides of an equation to any non-zero power.
• The rule for exponents when raising a power to a power is to multiply the exponents.
• Multiplying reciprocals always results in a 1.

All of this tells us that if we raise each side of the equation to the reciprocal of 4/3 power, we will end up with an exponent of 1 on the left side. The reciprocal of 4/3 is 3/4:

On the left side the exponent turns into a 1, as planned:

Simplifying the left side is simple. The right side is not so simple. If you have trouble with negative or fractional exponents I find it can be helpful to factor the exponent in a certain way:

1. If the exponent is negative, factor out a -1.
2. If the exponent is fractional and if the numerator is not a 1, then factor out the numerator. (You'll see what this means in a moment.)

The exponent on the right is not negative but it is a fraction whose numerator is not a 1. So we will factor out the 3:

Now, looking at the exponent in factored form, the 3 tells us we will be cubing and the 1/4 tells us that we will be finding a 4th root. And we can do these two operations in any order! Normally 4th roots aren't simple. However, since , finding the 4th root of 16 is rather easy. So we will start with the 4th root and then cube it:



x+1 = 8
Now we just subtract 1:
x = 7

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As often happens there are several ways this problem can be done. The most obvious first step is to use the rule for exponents when raising a power to a power. The rule is to multiply the exponents:

which simplifies as follows:


Next we try to simplify this expression even more. If you have trouble with negative and/or fractional exponents, I find it helpful to factor the exponent is a certain way:

1. If the exponent is negative, then factor out -1.
2. If the exponent is fractional and the numerator is not a 1, then factor out the numerator. For example, if the exponent was 2/3, then factor out the 2 as follows: 2*1/3.

After factoring the exponent in this way, look at the factors. Each one tells you one operation that will be done. And since multiplication is Commutative, these operations can be done in any order you choose!

Let's see how this works on your expression. First we factor the exponent:

Looking at the factors of the exponent, the -1 tells us that a reciprocal will be found. And the 1/4 tells us that a 4th root will be found. And the order we do them will not make any difference! So let's choose the order that seems the easiest. Normally a 4th root sounds difficult. But once you realize that 16 is then the 4th root is quite easy. So we will start with that. And then we will find a reciprocal.
(Since 1/2 is the reciprocal of 2)

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These are too many questions to post all at once. I will help with some of them in hopes that you will then know how to do the rest on your own.

First, let's be clear about what a logarithm is. Logarithms are exponents. The idea behind logarithms is that you can take any positive number (other than 1) and obtain any other positive number if you raise it to the correct power. For example, if you raise 3 to the correct power you can get a 4! This exponent is expressed as . In general represents the exponent to use on "a" to get a "b".

Once we get what a logarithm represents, some of your problems are extremely easy:
represents the exponent for 49 that results in 49. What power of 49 is 49? Answer: 1! (Or 1/1)
represents the exponent for 8 that results in 64. If you know your multiplication facts you know the answer is 2 (or 2/1).
represents the exponent for 1/125 that results in 125. If you know about negative exponents you will see that the exponent that turns 1/125 into 125 is -1! (-1/1)

Some you have to work at a little:
represents the exponent for 2 that results in 512. With the help of a calculator (or calculator computer program) and a little trial and error, you will find that 2 to the 9th power is 512.

And some are even harder:
represents the exponent for 16 that results in 512. But try as you might, you cannot find the right exponent. 512 is not a whole number power of 16. With these problems look to see if both the base and the argument are powers of some third number. In this case both the base, 16, and the argument, 512, are powers of 2! and . When this happens you use the base conversion formula, , to change the base:
So !

All of your other problems can be done in one of these ways:

• Just know what the right exponent is
• Trying different exponents on the base until you find the right one.
• If neither of the two above work, then see if both the base and argument are powers of the same third number. If they are then use the base conversion formula.

Final note: While all you problems can be done this way. Some logarithms cannot. The earlier example I gave you, cannot be found in any of these ways. This logarithm (and others like it) cannot be expressed as a rational number.

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First we can use a property of radicals, , to multiply in both the numerator and the denominator:

or

I left the radicand in the denominator in the form of for reasons that will become clear soon.
Next we use fractional exponents in the denominator to help us simplify further. Using the pattern we can rewrite the denominator as:

We can see that this fraction will reduce to:

Writing this back in radical form we get:

or

Now our fraction looks like:

Now we can use another property of radicals, to combine the two radicals into one:

which simplifies to:

And since , this cube root becomes a 3!

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This is all multiplication. So we can use the Commutative and Associative Properties to rearrange the order and grouping in any way we choose:

To multiply the radicals we use the property of all radicals: :

or

Just like fractions should be reduced, radicals should be simplified. Simplifying radicals involves finding factors of the radicand (the expression within a radical) that are powers of the type of root, if any. In this problem with its 4th root, we are looking for factors of 48 that are a power of 4. Since and since 16 is a factor of 48, this radical will simplify. We start by writing the radicand in factored form:

Then we use the property of radicals we used earlier. Only this time we use it in the other direction: to split a single radical into a product of radicals. By doing this we get the power of 4 factor into its own radical:

The 4th root of 16 is 2 so this becomes:

which simplifies to

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The solution provided by another tutor does not use the Quadratic Formula as the problem specifies and there is a typographics error in the first solution.

question 1 :
Dividing both sides by 2 will make the "a", "b" and "c" smaller. This will make the rest of the problem easier:

The Quadratic Formula:

which simplifies as follows:



In long form this is:
or
or
x = 14 or x = 9

question 2 :
One side needs to be zero so we start by subtracting 5 from each side:

The Quadratic Formula:

which simplifies as follows:




In long form this is:
or
or
x = 5 or x = -1

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The solution to the first problem provided by another tutor is correct. The solution to the second problem has an error.

2. the units digit of a two-digit number is 3 less than its tens digit. if the number is divided by the sum of its digits, the quotient is 6 and the remainder is 8. find the number.
The proper equations which express this are:
u = t-3


We can eliminate the fractions in the second equation if we multiply both sides of the equation by (t+u):

which simplifies as follows:
10t + u = 6*(t+u) + 8
10t + u = 6t + 6u + 8
Substituting t-3 for u we get:
10t + (t-3) = 6t + 6(t-3) + 8
which simplifies to:
11t - 3 = 6t + 6t - 18 + 8
11t - 3 = 12t - 10
Subtracting 11t from each side:
-3 = t - 10
Adding 10 to each side:
7 = t
This makes u = 7-3 or 4.
The original number is 74.

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First we use general exponential model equation and the initial information to find the specific exponential equation for this problem. In the general equation, the represents "the A when t = 0. The earliest value for A we have is from five years ago. So = 1400. This makes the other data, 1000 birds (A = 1000) five years later (t = 5) another point that should fit this equation. We now know 3 of the 4 unknowns in the general model. We can use the equation to find the 4th unknown, k:

First we divide by 1400:

which simplifies to

Next we find the natural log of each side:

Then we use a property of logarithms, , to move the exponent of the argument out in front. (It is this property that is the very reason we use logarithms on equations where the variable is in an exponent!)

By definition, ln(e) = 1 so this becomes:

Dividing by 5 we get:

This is an exact expression for k. This makes the specific equation to model this problem:


For practical use we should replace the k with its decimal approximation (which we can find using our calculator):



We can now use this equation to find when the bird population drops below 100:

We solve this for t, just like we solved the earlier equation for k:






39.3659175578729793 = t
So the bird population will drop below 99 approximately 39.4 years after the t=0 time. Since the t=0 time was 5 years ago, the bird population will drop below 100 approximately 34.4 years from now.

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Another tutor's solution is mostly correct. But there is a mistake.

Square both sides:


(Subtracting 2x from each side (this is where the mistake was made):

Factor:
(x-2)(x-8) = 0
From the Zero Product Property we know that this (or any) product can be zero only if one (or more) of the factors is zero. So:
x-2 = 0 or x-8 = 0
Solving these we get:
x = 2 or x = 8

Whenever you square both sides of an equation extraneous solutions may be introduced. Extraneous solutions are solutions that fit the squared equation but do not fit the original equation. These extraneous solutions can occur even if no mistakes have been made! For this reason you must check your answers any time you square both sides of an equation.

When checking, always us the original equation:

Checking x = 2:


Check failed! This is an extraneous solution and must be rejected.

Checking x = 8:


Check!

So the only solution to your equation is x = 8.

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I'm guessing that the equation is:

and not

If this is correct, then please use parentheses on multiple term exponents. For example:
2^(x+1) = 12
Tutors are more likely to help if the problem is stated clearly.

Solving equations where the variable is in an exponent usually involves logarithms. The base of the logarithm used does not make a significant difference. But if you want a decimal approximation (and you problem asks for a decimal), then you should choose a base of logarithm that your calculator "knows", like base 10 or base e (aka ln).

I'm going to use base e logarithms. (If you choose base 10 instead the answer works out the same!)

On the left side we can use a property of logarithms, , to move the exponent in the argument out in front. It is this very property that is the reason we use logarithms on problems like this. By moving the exponent, where the variable is, out in front the variable can now be solved for. Using this property of our equation we get:

Now we can solve for x. Dividing by ln(2) we get:

Subtracting 1 we get:

This is an exact expression for the answer. For the decimal approximation we go to our calculators to find the two logarithms:

x = 3.5849625007211562 - 1
which simplifies to:
x = 2.5849625007211562
Rounded to the nearest thousandth we get:
x = 2.585

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A ratio of common logs? A ratio of logs should bring to mind the base conversion formula: . So all we have to do is use this formula to convert the base 5.4 log into a ratio of common (i.e. base 10) logs:

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log(2x-1)=log(x+3)+ log(3)
Solving equations where teh variable is in the argument of a logarithm usually starts with transforming the equation into one of the following forms:
log(expression) = other-expression
or
log(expression) = log(other-expression)

Since your equation has nothing but logarithms the first form will more difficult to achieve. So we will aim for the second form. If we can find a way to combine the two logarithms on the right side into a single logarithm then we would have the second form. Those two logarithms are not like terms so we cannot just add them. But there is a property of logarithms, , which will allow us to combine the two logarithms on the right side:
log(2x-1)=log((x+3)*3)
which simplifies to:
log(2x-1)=log(3x+9)
We now have the equation in the second form. This equation says that the base 10 logarithm of 2x-1 is the same as the base 10 logarithm of 3x+9. The only way these two logarithms can be the same is if the arguments are the same. So:
2x-1 = 3x+9
This is a simple equation to solve. Subtracting 2x from each side we get:
-1 = x + 9
Subtracting 9 from each side and we get:
-10 = x

When solving logarithmic equations like this one, you must check your answers! You must ensure that any "solutions" do not make any argument (or base) of a logarithm negative or zero. This can happen even if not mistakes were made while solving.

If a "solution" does make an argument (or base) of a logarithm negative or zero then that solution must be rejected.

When checking, always use the original equation:
log(2x-1)=log(x+3)+ log(3)
Checking x = -10:
log(2(-10)-1)=log((-10)+3)+ log(3)
which simplifies to:
log(-20-1)=log(-7)+ log(3)
We can now see that an argument (actually two of them) is negative. So we must reject this solution. And since this was the only "solution" we found, your equation has no solution.

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If you have trouble with negative and/or fractional exponents, I find it helpful if you factor the exponent in a certain way:

• If the exponent is negative, factor out a -1.
• If the exponent is a fraction and the numerator is not a 1, factor out the numerator. (You'll undetsand this better shortly.)

Factoring your exponent this way we get:

Looking at the exponent's factors we see:

• A -1 which tells us that we will find a reciprocal.
• A 4 which tells us that we will be raising to the 4th power.
• A 1/3 which tells us that we will be finding a cube root.

And since multiplication is Commutative, we can do these three things in any order we choose! So we can shoose to do these things in the order that makes it easiest. (Note: What is easy for me is not necessarily easy for you. But don't worry. No matter what order you choose your answer will be the same as mine (as long as you do everything correctly) ) The order I choose is:

1. A cube root. Normally this is not something that would be easy. But since a cube root first looks easy.
2. Raise to the 4th power. The cube root in the first step will give us a fairly low number to raise to the fourth power.
3. A reciprocal. This will introduce a fraction and fractions make things a little more difficult so why not put it off until the end?

Here's the whole problem from the beginning to the end:

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Assuming the equation is:

(If this is correct, then please put parentheses around multiple term exponents like 2x-1.)

When the variable in an equation is in an exponent, logarithms are often used to solve the problem. (And I will solve this way.) But if you can make both sides of the equation powers of the same number, then there is a much easier way to solve the equation. Since 125 and 25 are both powers of 5, this easier solution is possible for this equation. I'll start with the quick solution.

First we rewrite each side as powers of 5. Since and since negative exponents are reciprocals, . So now we have:

On the right side, the rule for raising a power to a power is to multiply the exponents. This gibes us:

which simplifies to:

The only way for two powers of 5 to be equal is for the exponents to be equal. So:
-3 = 4x-2
Adding 2 we get:
-1 = 4x
Dividing by 4 we get:


If we were not able to rewrite both sides of the equation as powers of the same number (or if we didn't notice that we could) then we use logarithms to solve equations like this. The base of logarithm is not really important. But if you want a decimal approximation for the answer, choose a base of logarithm that your calculator "knows", like base 10 or base e (aka ln). I'll use base 10 logarithms:

Next we use a property of logarithms, , to move the exponent out in front as a coefficient. It is this very property that is the reason for using logarithms on equations like this. It allows us to move the exponent (where the variable is) out where we can solve for the variable. Using this property on the right side we get:

We can now solve for x. Dividing both sides by log(25) we get:

adding 1:

Dividing by 2:

Now we get out our calculators ans simplify the left side. You should get -0.25 (or some decimal extremely close to this). Since -0.25 = -1/4 this is the same answer we got the quick way.

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Since y is the debt and we are interested when the debt is 500 billion we will replace the y with 500 billion:


To solve an equation where the variable is in an exponent you usually isolate the base and its exponent, find the logarithm of each side and then solve the resulting equation. To isolate the base and its exponent we divide both sides by 1.21:

Next we find the logarithm of each side. The base of logarithm does not make a big difference. But if you want a decimal approximation for the answer it would be wise to choose a logarithm whose base your calculator "knows". lie base 10 or base e (aka ln). I'll use base 10:

Next we use a property of logarithms, , to move the exponent out in front as a coefficient. It is this very property that is the reason we use logarithms on problems like this. The property allows us to move the exponent (where the variable is) to a place where we can then solve for the variable.

To solve for x all we need to do now is to divide both sides by log(1.08):

This is an exact expression for the answer. For a decimal approximation we get out our calculators:


347.5427720429293512 = x
Since the United States did not exist in the year 347 this answer does not make sense. There are two reasons which may explain this answer:

• The debt in the equation was meant to be expressed in billions of dollars. If so then we should have used 500 and not 500000000000 for y. In this case our answer works out to be 78.2732100538809163 = x. This still doesn't make sense.either. However, ...
• These kinds of equations often use not the year itself but a number of years since some starting point year. You did not mention this in your post. But I think this is the best explanation for the answer we got. If "x" is the number of years since some starting point year, then we should add x to the starting point year to get the year when the debt was 500 billion.

My guess (and it is just a guess) is that we were supposed to use 500 and not 500000000000 for y and that the starting point year was 1900. This would make the answer to the problem: 1978.

In the future please post every detail of a problem.

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I solved this problem earlier. Click here to see the solution.

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The Law of Cosines has four variables: the lengths of the 3 sides of a triangle and the measure of one angle.

Your problem gives you 2 sides and the measure of one angle. So you know the values 3 of the 4 variables in the Law of Cosines. Anytime you know all but 1 variable in an equation you should be able to solve for the value of that last unknown. So we will use the Law of Cosines to solve this problem:

To convert BC into meters, we divide by 39.37:
BC = 423 inches = 432/39.37 meters = 10.7442214884429769 meters.
Also 132 degrees and 40 minutes = 132.67 degrees since 40 minutes is 2/3 of a degree and 2/3 as a decimal is approximately 0.67.
Inserting these values into our Law of Cosines equation we get:

To avoid confusion I am going to replace (AB) with an "x":

Simplifying we get:



This is a quadratic equation because of the . SO we want one side to be zero. Subtracting 67081 from each side we get:

Now we can use the Quadratic Formula:

which simplifies as follows:




In long form this is:
or
We can see that the second equation will give us a negative value. But the side of a triangle cannot be negative. So we will reject the second solution. The only solution we can use is:

Using a calculator to find the square root we get:


x = 251.4661081380912442
So AB is approximately 251.5 meters long.

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(To simplify matter I am going to use "x" instead of "theta". So just replace my x's with theta's.)

Since these points are on the unit circle, the x-coordinates are the cos of the angle to that point and the y-coordinates are the sin of the angle to that point. (This is so because the hypotenuse is always a 1 on the unit circle!) So:
cos(x) = 0.588 and sin(x) = 0.809

To find the coordinates of P(2x) we nned to find the cos(2x) and the sin(2x). Fortunately there are the double angle formulas which express these values in terms of cos(theta) and sin(theta):
cos(2x) = cos²(x) - sin²(x)
and
sin(2x) = 2sin(x)cos(x)
All we have to do is subsitute in the values for cos(x) and sin(x):
cos(2x) = (0.588)² - (0.809)²
and
sin(2x) = 2(0.809)(0.588)
I'll leave it up to you to simplify these expressions. The decimal you get for cos(2x) will be the desired x-coordinate and the decimal you get for sin(2x) will be the desired y-coordinate.

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The ratio for tangent is opposite over adjacent. Sketching this triangle may help you understand how to do this problem.

Since AB is the longest side it must be the hypotenuse. This makes angle C the right angle. So draw a triangle with C being a right angle and with the sides being the lengths given.

Looking at this triangle you should be able to see that the opposote side for angle B is AC (or 12) and the adjacent side for angle B is BC (or 5) So this makes:
tan(B) = 5/12

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Lines have infinite length, including the line that passes through A and B.

But if the problem is to find the length of the line segment whose endpoints are A and B, then we can use the distance formula, :

which simplifies as follows:



So the distance between A and B (or the length of line segment AB) is .

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I'd be curious to know how you can solve for x without knowing how to write this equation in exponential form.

Solving logarithmic equations where the variable is in the argument (or base) of a logarithm usually starts with transforming the equation into one of the following forms:
log(expression) = other-expression
or
log(expression) = log(other-expression)

With the "non-log" term of 8 it will be more difficult to reach the second form. So we will aim for the first form. We want one side of the equation to be a single logarithm. For this we can use the property of logarithms: . Using this on your equation we get:

which simplifies to:


We now have the first form. The next step with the first form is to rewrite the equation in exponential form. In general } is equivalent to . Using this pattern on your equation we get:
(since the base of "log" is 10)
which simplifies to:

This equation we can solve. Dividing both sides by 3 we get:

Now we find the square root of each side:

(Note: Algebra.com's software will not let me use the "plus or minus" symbol without something in front of it. This is why the extra zero is there in front.)
In long form this is:
or

With any logarithmic equation like yours you must check your answers. You must ensure that no arguments (or bases) of any logarithms become negative or zero. This can happen even if no mistakes were made! So even if you make no errors, you still have to check for this. And if you find that an "answer" does make an argument (or base) negative or zero then you must reject that "answer".

Always use the original equation to check:

Checking :

We can see already that both arguments will be positive. So there is no reason to reject this solution. We have done the required part of the check for this solution. The rest of the check will just tell us if we made an error. You are welcome to finish the check if you like.

Checking :

We can already see that both arguments are going to be negative, So we must reject this solution. (If even just one argument had been negative or zero we would still reject the solution.

So the only solution to you equation is: . This square root is not in proper form. (There should not be a fraction in a square root. So we should rationalize the denominator:




And then simplify the numerator:



This is the proper form for your solution.

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What is "to the 5th power?
The x?
Or the entire fraction, 5/x?

Please use parentheses to make your problems clear. Also use "^" to indicate powers. For example:
5/(x^5)
or
(5/x)^5

Tutors are more likely to help when the problems are clearly stated.

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A number in scientific notation is of the form:
a x
where
and n is an integer.
Note 1: The "a" can be equal to 1. But it cannot be 10.
Note 2: The "x" is a multiplication sign. Scientific notation often uses the old-fashioned "x" for multiplication.

To change a number into scientific notation:

1. Write the number in decimal form. If the number is a fraction, divide it. If the number is an integer, put a decimal at the end.
2. Move the decimal so that is is between the leftmost non-zero digit and the digit to its right. This new number will be the "a".
3. The number of places you moved the decimal point will be . If you moved the decimal to the left, n will be positive and if you moved the decimal to the right then the exponent will be negative.

Let's try this on your number:
1) Write with a decimal. Your number is already in decimal form.
2) Move the decimal. The leftmost non-zero digit is the 6. So we move the decimal between the 6 and the digit to its right, the 8:
6.8
This is the "a"
3) We moved the decimal point to the right above so the exponent, n, will be negative. And we moved it 7 places so the exponent is -7

So 0.00000068, in scientific notation is 6.8 x .

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I assume you mean the 12th root of 64:

When simplifying radicals the first thing you look for are factors of the radicand (the expression inside a radical) that are powers of the type of root. In this case we would look for factors of 64 that are powers of 12. Since factoring out a 1 is rarely useful we start by checking 2. . And obviously higher numbers to the 12th power will be even higher. So there are no factors of 64 that are 12th powers of anything (except 1).

Another way to simplify radicals can be done once you've learned about fractional exponents. If the radicand is a power of some sumber, write it that way. Since we have 3 choices as to how we rewrite 64. In this case choose an exponent that is a factor of the root. But all these exponents, 2, 3 and 6, are all factors of 12. In this case choose the smallest base. So we will use :

Next we rewrite the radical with a fractional exponent:

This fraction reduces. (This is why we look for an exponent that is a factor of the type of root.)

Now we can rewrite it back in radical form:

or

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What is "to the 3rd power"?
The b?
The 9ab?
The square root of 9ab?

Please use parentheses to make your problem clear. Tutors are more likely to help when the problem is clearly stated.

Also, use "^3" instead of "to the 3rd power".

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3ln(5)+2ln(x)
These are not like terms so we cannot simply add them. But there is a property, , which can be used to combine two logarithms into one. This property requires all of the following:

• There is a "+" between the two logarithms. (Note: there is also another property for when there is a "-" between two logarithms.)
• The bases of the two logarithms must be the same.
• The coefficients of the two logarithms must be 1's.

Your expression meets the first two requirements but not the third one. Fortunately there is another property of logarithms, , which allows us to move a coefficient of a logarithm into its argument as an exponent. Using this property on both of your logarithms we get:

which simplifies to:

Now that the coefficients are 1's we can use the other property to combine the two logarithms:

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In order to determine whether this is true or false, we are going to manipulate the left side to see if we can make it look like the right side. But before we start, I am going to replace with . (Remember fractional exponents?):

Since the right side does not have a 3 in front of the logarithm I will start by using the property of logarithms, , to move the coefficient into the argument as an exponent:

which simplifies to:

Since is clearly not the same as "a" we can see that this (and the original equation) is not a true statement.