Question

Answers................ ​

Answer 1

`\\ \rm\Rrightarrow log_2(x^2-x+2)=1+2log_2x`

`\\ \rm\Rrightarrow log_2(x^2-x+2)=log_22+2log_2x`

`\\ \rm\Rrightarrow log_2(x^2-x+2)=2log_2(2x)`

`\\ \rm\Rrightarrow log_2(x^2-x+2)=log_2(2x^2)`

`\\ \rm\Rrightarrow x^2-x+2=2x^2`

`\\ \rm\Rrightarrow x^2+x-2=0`

`\\ \rm\Rrightarrow (x+2)(x-1)=0`

• x=-2,1

log is always positive so x=1

Laws used

• log_a(a)=1
• loga^m=mloga
• log(ab)=loga+logb

Answer 2

x=1

Explanation:

log2( x^2 -x+2) = 1+2log2(x)

Rewriting 1 as log2(2)

log2( x^2 -x+2) = log2(2)+2log2(x)

We know that a log b = log a^b

log2( x^2 -x+2) = log2(2)+log2(x^2)

we know log a + log b = log (ab)

log2( x^2 -x+2) = log2(2*x^2)

Since the bases are the same the terms inside must be equal

x^2 -x+2 = 2x^2

Subtract 2x^2 from each side

-x^2 -x+2 = 0

Multiply by -1

x^2 +x-2 = 0

Factor

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

Using the zero product property

x+2 = 0 x-1=0

x=-2 x=1

Checking the solutions

log2( x^2 -x+2) = 1+2log2(x)

X cannot be negative because 2 log2(x) cannot be negative

log2( 1^2 -1+2) = 1+2log2(1)

x=1