# Find x for lg2 + lg[ 4^(x-2) + 9 ]= 1 + lg[ 2^(x-2)+ 1 ]

hala718 | High School Teacher | (Level 1) Educator Emeritus

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lg2 + lg[4^(x-2) + 9] = 1+ lg[2^(x-2) + 1]

we know that log 10 = 1

We know that log a + log b = log a*b

==> log 2*[4^(x-2) + 9] =  log 10*[2^(x-2)+1]

We know that if log a = log b, ==> a = b

==> 2*[4^(x-2) + 9] = 10*[2^)x-2) + 1]

==> 2*4^(x-2) + 18 = 10*(2^(x-2) + 10

==> 2*4^(x-2) + 8 = 10*2^(x-2)

==> 2*[2^(x-2)]^2 = 10*2^(x-2) + 8 = 0

Assume that 2^(x-2) = y

==> 2y^2 - 10y + 8 = 0

==> y^2 - 5y + 4 = 0

==> (y-4)(y-1) = 0

==> y1= 4  ==> 2^(x-2) = 4 ==> x-2 = 2 ==> x1 = 4

==> y2= 1 ==> 2^(x-2) = 1 ==> x-2 = 0 ==> x2= 2

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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We'll determine the constraints of existance of logarithm function.

4^(x-2) + 9 > 0

We'll subtract 9 both sides:

4^(x-2) >  9

For any value of x, 4^(x-2) > 0.

2^(x-2)+ 1 > 0

We'll subtract 1 both sides:

2^(x-2)>-1

For any value of x, 2^(x-2) > 0.

Now, we'll solve the equation.

We notice that, to the left side, we have the sum of 2 logarithms that have the same base. We'll use the product property of logarithms (the sum of logarithms is the logarithm of the product).

lg2 + lg[ 4^(x-2) + 9 ]=lg {2*[ 4^(x-2) + 9 ]}

We'll re-write the value 1, from the right side, using the decimal logarithm.

1 = lg 10

We'll re-write the right side:

1 + lg[ 2^(x-2)+ 1] = lg 10 + lg[ 2^(x-2)+ 1]

We'll also transform the sum into a product:

lg 10 + lg[ 2^(x-2)+ 1] = lg 10*[ 2^(x-2)+ 1]

We'll re-write the entire equation:

lg {2*[ 4^(x-2) + 9 ]} = lg {10*[ 2^(x-2)+ 1]}

Since the bases are matching, we'll apply the one to one property:

2*[ 4^(x-2) + 9] = 10*[ 2^(x-2)+ 1]

We'll remove the brackets:

2*4^(x-2) + 18 = 10*2^(x-2) + 10

We'll write 4^(x-2) as 4^x / 4^2, 2^(x-2) as 2^x / 2^2 and we'll subtract 18 both sides:

2 * 4^x / 4^2 = 10*2^x / 2^2 + 10 - 18

4^x / 2^3 = 5*2^x / 2 - 8

We'll move all terms to one side:

4^x / 8 - 5*2^x / 2 + 8 = 0

We'll bring all ratios to the common denominator, 8.

4^x - 5*4*2^x + 64 = 0

We'll note 2^x = t.

t^2 - 20t + 64 = 0

t1 = [20 + sqrt(400 - 256)] / 2

t1 = (20+12)/2

t1 = 32/2

t1 = 16

t2 = (20-12)/2

t2 = 8/2

t2 = 4

2^x = t1

2^x = 16

2^x = 2^4

x = 4

2^x = t2

2^x = 4

2^x = 2^2

x = 2

Since both solutions are positive, they are both valid.

neela | High School Teacher | (Level 3) Valedictorian

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lg2+lg(4^(x-2) +9] = 1+lg(2^(x-2)+1] to solve for x.

Solution:

We use lga +lgb = lgab  and

1 = lg10.

so the given equation  becomes:

lg { 2[4^(x-2) + 9]} = lg10 + lg{2^(x-2) +1}

lg{ 2{4^(x-2)+9]} = lg { 10 [ 2^(x-2)+1}.

Take anti logarithms:

2{4^(x-2)+9} = 10{ 2^x-2) +1}

(2*4^x)/4^2 + 18 = 10 * 2^x /2^2 + 10

4^x )/8 + 18 = 5(2^x )/2 +10. Multiply by 8.

4^x + 144 = 20(2^x ) +80.

(2^x)^2 +144 = 20(2^x) +80 , as 4^x = (2^2)^x = (2^x)^2.

t^2 +144 = 20x+80 , where t = 2^x.

t^2-20x +144-80 = 0

t^2-20x+64 = 0

(t-16)(t-4) = 0

t-16 = 0 , t-4 = 0

t = 2^x =16 , x = 4

t = 2^x = 4, x =2.

So x=2 Or x=4.