# Maths SURDS and INDICES problems?

+1 vote
Solve

1)(√3)x = x + √3

2)2+(√3) x= x/√3

3) (1/√2)+x= x/√2

4) (27x^-3)/2 = 1

5) (2x^3) - 128 =0

6) (32x^-3) = (√x)/4

7) 5 * 5^x = 25 ^x^2

+1 vote

5)       (2x^3) - 128 =0

(2x^3)-128+128=0+128

(2x^3)=128

Divide each side by 2

(2x^3)/2=128/2

x^3=64

x^3=4^3

x=4

So,the value of x is 4

Good approach

1 ) (√3)x = x + √3

Subtract x from each side

(√3)x - x = x + √3 - x

(√3)x - x = √3

Using distributive property ab - ac = a (b - c)

x (√3 - 1) = √3

x = √3 /√3 - 1

Get rid of the denominator surd we multiply √3 /√3 - 1 by √3 + 1 /√3 + 1 like so.

x = √3 /√3 - 1 ×√3 + 1 /√3 + 1

x = √3 (√3 + 1) /(√3 + 1)(√3 - 1)

x = √3 (√3 + 1) /(3 - 1)

x = √3 (√3 + 1) / - 2

x = - √3 (√3 + 1) /  2

x = √3 (√3 + 1) /(3 - 1)

x = (√3√3 + √3) /(2)

x = (3 + √3) /(2)

The solution x = (3 + √3)/2.

2 ) 2+(√3) x= x/√3

Subtract √3x from each side

2+√3x - √3x= x/√3 - √3x

x/√3 - √3x = 2

Using distributive property ab - ac = a ( b - c )

x( 1/√3 -√3) = 2

x( 1 -3 /√3) = 2

x( -2 /√3) = 2

Divide each side by 2

-x / √3 = 1

x = - √3

6 ) (32x^-3) = (√x)/4

Multiply each side by 4x^3

4x^3 * 32x^-3 =4x^3 * (√x)/4

128 = √x * x^3

128 = x^7/2

2^7 = x^7/2

Take the seventh root of each side to get (√x) = 2

Squaring each side

x = 4

7)  5*5^x = 25 ^x^2

25^x = 25^x^2

Taking log base 25 of each side

x=x^2

Subtract x from each side

x - x = x^2 - x

x^2 - x = 0

x ( x- 1) = 0

x = 0 or x - 1 = 0

x = 0 or x = 1

The solution is x = 1 or -1/2.
3) (1/√2)+x= x/√2

given (1/√2)+x=x/√2

subtrct from x each side

1/√2+x-x=x/√2-x

1/√2=x(1/√2-1)

1/√2=x(1-√2)/√2

multiply each side by √2

√2/√2=x(1-√2)/√2*√2

1=x(1-√2)

x=1/1-√2

x=(1/1-√2)*1+√2/1+√2

x=1+√2/1-2

x= -(1+√2)

1 ) (√3)x = x + √3

Subtract x from each side

(√3)x - x = √3

Using distributive property ab - ac = a (b - c)

x (√3 - 1) = √3

x = √3 /(√3 - 1)

Get rid of the denominator surd multiplying √3 /(√3 - 1) by (√3 + 1) /(√3 + 1)

x = √3 /(√3 - 1) ×(√3 + 1) /(√3 + 1)

x = √3 (√3 + 1) /(√3 + 1)(√3 - 1)

x = √3 (√3 + 1) /(3 - 1)

x = √3 (√3 + 1) / 2

x = √3 (√3 + 1) /  2

5* 5x = 25 ^2

5^(1+x) = 5^2x^2

Taking log base 5 of each side

1+x=2x^2

2x^2-x-1=0

2x^2-2x+x-1=0

2x(x-1)+1(x-1)=0

(x-1)(2x+1)=0

x-1=0 , 2x+1=0

x=1,x=-1/2

) (√3)x = x + √3

Subtract x from each side

(√3)x - x = x + √3 - x

(√3)x - x = √3

Using distributive property ab - ac = a (b - c)

x (√3 - 1) = √3

x = √3 /√3 - 1

Get rid of the denominator surd we multiply √3 /√3 - 1 by √3 + 1 /√3 + 1 like so.

x = √3 /√3 - 1 ×√3 + 1 /√3 + 1

x = √3 (√3 + 1) /(√3 + 1)(√3 - 1)

x = √3 (√3 + 1) /(3 - 1)

x = 3+√3/2

7) The equation is 5 * 5x = 25x^2.

51 * 5x = 25x^2

Apply Product of Powers Property: am * an = am+n.

51+x = 25x^2

Write 25 as 52.

51+x = 52x^2

Since bases are equal, equate the exponents.

1 + x = 2x2

2x2- x - 1 = 0

2x2 - 2x + x - 1 = 0

2x(x - 1) + 1(x - 1) = 0

(2x + 1) (x - 1) = 0

2x + 1 = 0 or x - 1 = 0

Take 2x + 1 = 0

2x = -1

x = -1/2

Take x - 1 = 0