can someone please help me with these mat log problems?

Question

solve for x

1. squareroot 8x-7=x+1

2. x^2-4x-12>or equal to 0

3. x^2/3+5x^1/3-14=0

solving logs

4. y=log8 2

5. -3=lOg5 x

6. logb 3=1/4

thanks for help I'm now learning please help

Answer 1

Teeks suggests the best strategy: log b (arg) = y converts to an equivalent exponent expression:

b^y = arg

In the exponent equation you either find each side has the same (prime usually) base, OR you force each side to use the same base (10 or e).

prob 5: 5^(-3) = x then x = 1/(5^3) = 1/125

prob 6: b^(1/4) = 3 then (raising each side to the 4th power: b = 3^4 or 81

Answer 2

Marcella did a good job with 1 and 2, so I'll take the rest.

#3 is a quadratic in disguise. Look at the exponents on x: they are 2/3 and 1/3. The key is that one of them is double the other. We can unravel this problem by using another variable to stand for the smaller power of x.

Say, t=x^1/3.
Square both sides of this to get t^2=x^2/3.

Now we can replace x^2/3 and x^1/3 with t^2 and t:
t^2 +5t - 14=0.

Factor: (t+7)(t-2)=0.

Now t=-7 or t=2.

But we weren't trying to find t; we wanted x. We have to reverse our substitution. If t=x^1/3, then x=t^3. So, x=(-7)^3=-343, or else x=2^3=8.

Problems 4-6 are very similar to each other. A useful thing to know is that logarithmic equations (like the ones you wrote) can be translated into exponential equations. To do this, simply move the base of the logarithm to the other side of the equation, where it becomes the base of an exponential, and erase the word "log". For instance:

4. y=log8 2. Here 8 is the base of the logarithm, so we move it across the "equals" sign and use it as the base of an exponential: 8^y. Meanwhile, we erase the word "log" on the right side. In other words, we get 8^y=2. The original equation was in "log form"; the new one is in "exponent form". They are different ways of saying the same thing.

To finish, we need to know how to handle exponents. Since 8=2^3, we have
8^y=(2^3)^y=2^(3y). Meanwhile, 2=2^1. This gives us 2^(3y)=2^1, or 3y=1. So, y=1/3.


5. First we translate the log equation -3 = log5 x into exponent form by moving the base (5) to the other side: 5^(-3)=x. Now we need to know how to simplify that power. 5^(-3) = 1/(5^3) = 1/125 or 0.008.


6. Translate the log equation logb 3 = 1/4 into exponent form by moving the base (b) to the other side: 3=b^(1/4). Raising both sides to the 4th power, we find b=3^4=81.

It's also good to know how to reverse the above method, and write an exponential equation in log form. It's easy, but there's some fine print.

Example: 5=2^x is written in log form by moving the base of the exponent to the opposite side, and writing in "log" (using that base). That is, we move the 2 across. We get log2 5 = x.

The fine print is that this does NOT work if anything is being added or multiplied onto the exponential term. For example, if we have

15=3(2^x)+3, we CANNOT grab that 2 out. We have to rearrange the equation first, to get the exponential term alone:

15 = 3(10^x) + 6
(subtract 6)
9 = 3(10^x)
(divide by 3)
3 = 10^x Now the exponential term is by itself, with no extra coefficients or terms. It is now safe to translate into log form:
log10 3 = x.

log10 is the one your calculator knows as LOG, so you could compute this if you wanted.

Answer 3

1. sqrt(8x - 7) = x + 1
8x - 7 = x² + 2x + 1
x² - 6x + 8 = 0
(x - 2)(x - 4) = 0
x = 2 or 4

2. x² - 4x - 12 >= 0
(x - 6)(x + 2) >= 0
Parabola opens upwards; zeros at x = -2 and 6.
x <= -2 or x >= 6

3. x^(2/3) + 5x^(1/3) - 14 = 0
u = x^(1/3)
u² + 5u - 14 = 0
(u + 7)(u - 2) = 0
u = -7 or 2
x = (-7)^3 or 2^3
x = -343 or 8

4. y = log(base 8) 2
While the method suggested by teeks will work, I like the change of base formula much better. The change of base formula states:

log(base B) N = log(base C) N / log(base C) B

C can be anything you choose. This means that you can easily calculate logs on your calculator using the natural log or log base 10 functions: log(base 8) 2 = ln 2/ln 8.
Use change of base formula:
y = ln(2)/ln(8)
y = 1/3

5. -3 = log(base 5) x
(-3)^5 = x
x = (-3)(-3)(-3)(-3)(-3)
x = -243

6. log(base b) 3 = 1/4
Use change of base formula:
ln 3/ln b = 1/4
4 ln 3 = ln b
b = exp(4 ln 3)
b = 81

Answer 4

Log problem #4:

y = log(base 8) 2

To solve this you want to make it into a regular exponential equation. In both logarithms and exponential equations, the base is the same. The answer to a log is the power you raise the base to.

Therefore, the logarithm above turns into:

8^y = 2.

To solve this you want to make the bases on both sides of the equation the same. Do you see that 8 = 2^3 and 2 = 2^1?

(2^3)^y = 2^1

Use your power of a power rule.

2^(3y) = 2^1.

You should have learned the property that a^b = a^c only if b=c. Use this to solve.

3y = 1
y = 1/3

Hope that helps!

Answer 5

1. sqrt(8x-7) = x + 1
Square both sides: 8x - 7 = x^2 + 2x + 1
get everything on one side: 0 = x^2 - 6x + 8
factor: 0 = (x - 2)(x -4)
x = 2 or 4

2. x^2 - 4x - 12 > 0
Factor: (x-6)(x+2) > 0

So both factors have to be positive or both factors have to be negative to satisfy the inequality. If both are positive, then x>6. If both are negative, then x < -2. So the solution is: x > 6 or x < -2.

Answer 6

1. x= 2 or x=4
2. x= 6
3. x= 4.48 or x=-9.44
4. y= 1/3
5. x= 4.29