Math help please!!! need soon?

Question

Solve these math problems please!

1. (sqrt X+1) = (sqrt x+6) -1

2. 8^2x+3=4*2^x+1

3. (x+1)/(3x-6)= (5x)/(6) + (1)/(x-2)

4.(1/625)^(-3/4)

5. e^2

6. (3- (sqrt 2))/ (2sqrt3) +5)

if you can figure it out

(x^2+ax) over(2a-x) * (x^2-3ax+2a^2)over(a^2-x^2)divided by (x+2a) over (x+a)

Answer 1

(Problem 1)

√(x + 1) = √(x + 6) - 1

Square both sides:

[ √(x + 1) ]² = [ √(x + 6) - 1 ]²

x + 1 = √(x + 6)² - 2·1·√(x + 6) + 1²

x + 1 = (x + 6) - 2·√(x + 6) + 1

x + 1 = x + 7 - 2·√(x + 6)

1 = 7 - 2·√(x + 6)

-6 = -2·√(x + 6)

3 = √(x + 6)

Square again:

3² = [ √(x + 6) ]²

9 = x + 6

3 = x

Since you squared, check to make sure this solution works in the original problem. It does, so you have one solution:

x = 3

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(Problem 2)

I think you left out some important parenthesis here...

8^(2·x + 3) = 4·2^(x + 1)

Method 1:

When you multiply two same bases together, you add their exponents, so (2^x)(2) = 2^(x+1) and vice versa:

8^(2·x)·8³ = 4·2^(x)·2¹

8³·8^(2·x) = 8·2^(x)

8²·8^(2·x) = 2^(x)

64·8^(2·x) = 2^(x)

8 is a power of two:

64·(2³)^(2·x) = 2^(x)

64·2^(6·x) = 2^(x)

Combine the bases of two:

64 = 2^(x)/2^(6·x)

64 = 2^( x - 6·x )

64 = 2^( -5·x )

Take log base two of both sides to cancel the base of 2:

log,,[ 64 ] = log,,[ 2^( -5·x ) ]

6 = -5·x

-6/5 = x

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Method 2:

8^(2·x + 3) = 4·2^(x + 1)

Take log base 2 right away:

log,,[ 8^(2·x + 3) ] = log,,[ 4·2^(x + 1) ]

Use the properties of log:

log,,[ 8^(2·x + 3) ] = log,,[ 4 ] + log,,[ 2^(x + 1) ]

(2·x + 3)·log,,[ 8 ] = log,,[ 4 ] + (x + 1)·log,,[ 2 ]

Evaluate the logs:

(2·x + 3)·( 3 ) = ( 2 ) + (x + 1)·( 1 )

Solve:

6·x + 9 = x + 3

6·x - x = 3 - 9

5·x = -6

x = -6/5

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(Problem 3)

( x + 1 ) / ( 3·x - 6 ) = 5·x / 6 + 1 / ( x - 2 )

Combine the fractions on the right by getting a common denominator:

( x + 1 ) / ( 3·x - 6 ) = 5·x·( x - 2 ) / [ 6·( x - 2 ) ] + 6 / [ 6·( x - 2 ) ]

( x + 1 ) / ( 3·x - 6 ) = [ 5·x·( x - 2 ) + 6 ] / [ 6·( x - 2 ) ]

( x + 1 ) / [ 3·( x - 2 ) ] = [ 5·x·( x - 2 ) + 6 ] / [ 6·( x - 2 ) ]

Cross-multiply:

( x + 1 )·[ 6·( x - 2 ) ] = ( 5·x² - 10·x + 6 )·[ 3·( x - 2 ) ]

Cancel any common factors before expanding:

( x + 1 )·6 = ( 5·x² - 10·x + 6 )·3

( x + 1 )·2 = ( 5·x² - 10·x + 6 )

2·x + 2 = 5·x² - 10·x + 6

0 = 5·x² - 12·x + 4

0 = ( x - 2)·( 5·x - 2 )

x = 2 or x = 2/5

Check these in the original problem. The solution x = 2 causes the denominator to be zero in at least one place, 1/(x-2). So 2 is not a valid solution.

Answer:

x = 2/5

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(Problem 4)

(1/625)^(-¾)

Negative powers can flip what it is applied to (i.e. gets the reciprocal) and become positive:

= (625/1)^(¾)

= (625)^(¾)

Fractional powers get roots. The fourth root of 625 is 5.

= (5^4)^(¾)

= (5)^(3)

= 125

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(Problem 5)

e² cannot be simplified. It is an irrational number.

It can be approximated. e is approximated by:

e ≈ 2.7182818284590452353602874713527

Thus, e² is approximated by:

e² ≈ 7.389056098930650227230427460575

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(Problem 6)

This cannot be solved, but the denominator can be rationalized (meaning it won't have the irrational, square root part):

[ 3 - √(2) ] / [ 2·√(3) + 5 ]

Multiply by the conjugate of the denominator:

= [ 3 - √(2) ]·[ 2·√(3) - 5 ] / { [ 2·√(3) + 5 ]·[ 2·√(3) - 5 ] }

= [ 3·2·√(3) + 3·(-5) - √(2)·2·√(3) - 5· -√(2) ] / [ 2²·√(3)² - 5² ]

= [ 6·√(3) - 15 - 2·√(6) + 5·√(2) ] / ( 4·3 - 25 )

= [ 6·√(3) - 15 - 2·√(6) + 5·√(2) ] / -13

= [ 2·√(6) - 6·√(3) - 5·√(2) + 15] / 13

You could split that into separate fractions, but they would all be over 13 anyhow.

= 2·√(6)/13 - 6·√(3)/13 - 5·√(2)/13 + 15/13

Answer 2

Were you actually asked to solve these?  Or use some kind of approximation method?  Because the first two don't look solvable, and the last three are just expressions.

Answer 3

4. 125

All I know is number 4...srry...