Question

Lg(4^x+9)=lg5+lg(2^x+1 )

Answer 1

Lg(4^x +9)=lg[5(2^x +1)]
2^2x -5*2^x+9-5=0
2^2x -5*2^x+4=02^x=a
a²-5a+4=0
a1+a2=5 U a1*a2=4
a1=1⇒2^x=1⇒x=0
a2=4⇒2^x=4⇒x=2

Answer 2

To solve the equation lg(4^x+9) = lg(5) + lg(2^x+1), you can use logarithm properties.

1. Start by simplifying the equation using logarithm properties:

   lg(4^x+9) = lg(5) + lg(2^x+1)

2. Since both sides have a common logarithm base (base 10), you can eliminate the logarithms by setting the expressions inside the logarithms equal to each other:

   4^x + 9 = 5 * (2^x + 1)

3. Now, let's isolate the exponential terms. First, distribute 5 on the right side:

   4^x + 9 = 5 * 2^x + 5

4. Move the 5 * 2^x term to the left side of the equation:

   4^x - 5 * 2^x + 9 = 5

5. To make it easier to solve for x, let's make a substitution. Let y = 2^x. Now, the equation becomes:

   (y^2) - 5y + 9 = 5

6. Rearrange the equation:

   y^2 - 5y + 9 - 5 = 0

7. Simplify further:

   y^2 - 5y + 4 = 0

8. Now, you have a quadratic equation. Factor it:

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

9. Solve for y:

   y = 4 or y = 1

10. Remember that we made a substitution earlier: y = 2^x. Now, substitute back:

2^x = 4 or 2^x = 1

11. Solve for x:

For 2^x = 4, take the logarithm of both sides (base 2):

x * log2(2) = log2(4) x = 2

For 2^x = 1, we know that any number raised to the power of 0 is 1:

x = 0

So, the equation has two solutions: x = 0 and x = 2.