Chapter ISG, Problem 7.3.4P

a.

To determine

To justify the given statement.

Answer to Problem 7.3.4P

J’s equation is not equal to $8^{(2+x)}={16}^{(2.5-0.5x)}$ .

V’s equation is not equal to $8^{(2+x)}={16}^{(2.5-0.5x)}$ .

Explanation of Solution

Given information:Equations given:

J	V
$2^{(8+4x)}=2^{(10-2x)}$	$4^{(3+1.5x)}=4^{(5-x)}$

  $8^{(2+x)}={16}^{(2.5-0.5x)}$

Formula used:

If $x^a=x^b$ ,

Then $a=b$ ,

Therefore, If bases are same than powers are also equal.

Calculation:

On solving the given equation, $8^{(2+x)}={16}^{(2.5-0.5x)}$

8 can be written as $2^3$ and 16 can be written as $2^4$ .

On solving the equation,

  ${(2^3)}^{(2+x)}={(2^4)}^{(2.5-0.5x)}$

Opening the brackets.

  $2^3{}^{(2+x)}=2^4{}^{(2.5-0.5x)}$

If bases are same than powers are also equal.

Therefore,

  $3(2+x)=4(2.5-0.5x)$

Opening the brackets.

  $6+3x=10-2x$

Now, putting constants on one side,

  $6-10=-2x-3x$

On solving further,

  $-4=-5x$ ,

So,

  $\frac{-4}{-5}=x$

Canceling negative signs,

  $x=\frac{4}{5}$

Value of is $x=\frac{4}{5}$ .

Substituting the value of x in all 3 equations,

  $8^{(2+x)}={16}^{(2.5-0.5x)}$

8 can be written as $2^3$ and 16 can be written as $2^4$ .

  ${(2^3)}^{(2+x)}={(2^4)}^{(2.5-0.5x)}$

Substituting the value of x

  $2^3{}^{(2+\frac{4}{5})}=2^4{}^{(2.5-0.5(\frac{4}{5}))}$

Solving inside the bracket

  $2^{3(\frac{10+4}{5})}=2^{4(2.5-\frac{2}{5})}$

Again,

  $2^{3(\frac{14}{5})}=2^{4(\frac{12.5-2}{5})}$

So,

  $2^{\frac{42}{5}}=2^{4(\frac{10.5}{5})}$

Opening the brackets,

  $2^{\frac{42}{5}}=2^{\frac{42}{5}}$

Substituting the value of x in $2^{(8+4x)}=2^{(10-2x)}$

  $2^{(8+4(\frac{4}{5}))}=2^{(10-2(\frac{4}{5}))}$

Solving inside the bracket

  $2^{(8+\frac{16}{5})}=2^{(10-\frac{8}{5})}$

Again,

  $2^{(\frac{40+16}{5})}=2^{(\frac{50-8}{5})}$

So,

  $2^{(\frac{56}{5})}=2^{(\frac{42}{5})}$

Therefore left and side is not equals to right hand side.

So, J’s equation is not equal to $8^{(2+x)}={16}^{(2.5-0.5x)}$ .

Substituting the value of x in $4^{(3+1.5x)}=4^{(5-x)}$

4 can be written as $2^2$ so,

  $2^{2(3+1.5x)}=2^{2(5-x)}$

Substituting the value of x

  $2^{2(3+1.5(\frac{4}{5}))}=2^{2(5-\frac{4}{5})}$

Solving inside the bracket

  $2^{2(3+\frac{6}{5})}=2^{2(5-\frac{4}{5})}$

Taking least common factor,

  $2^{2(\frac{15+6}{5})}=2^{2(\frac{25-4}{5})}$

After solving

  $2^{2(\frac{21}{5})}=2^{2(\frac{21}{5})}$

So,

  $2^{\frac{42}{5}}=2^{\frac{42}{5}}$

Therefore left and side is equals to right hand side.

So,V’s equation is not equal to $8^{(2+x)}={16}^{(2.5-0.5x)}$ .

b.

To determine

To calculate the value of x.

Answer to Problem 7.3.4P

Value of $x=\frac{4}{5}$ .

Explanation of Solution

Given information:

  $8^{(2+x)}={16}^{(2.5-0.5x)}$

Formula used:

If $x^a=x^b$ ,

Then $a=b$ ,

Therefore, If bases are same than powers are also equal.

Calculation:

On solving the given equation,

8 can be written as $2^3$ and 16 can be written as $2^4$ .

On solving the equation,

  ${(2^3)}^{(2+x)}={(2^4)}^{(2.5-0.5x)}$

Opening the brackets.

  $2^3{}^{(2+x)}=2^4{}^{(2.5-0.5x)}$

If bases are same than powers are also equal.

Therefore,

  $3(2+x)=4(2.5-0.5x)$

Opening the brackets.

  $6+3x=10-2x$

Now, putting constants on one side,

  $6-10=-2x-3x$

On solving further,

  $-4=-5x$ ,

  $\frac{-4}{-5}=x$

Canceling negative signs,

  $x=\frac{4}{5}$

Value of $x=\frac{4}{5}$ .