Question

Log2(lgx+2√lgx+1)-2log4(√lgx+1)=1

Answer 1

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Answer 2

To solve the given equation, let's start by simplifying it step by step. I'll assume that the base of the logarithms is 2 unless specified otherwise.

Given equation:

log2(lgx + 2√lgx + 1) - 2log4(√lgx + 1) = 1

Step 1: Convert everything to a common base, which is 2 in this case.

Using the change of base formula: loga(b) = logc(b) / logc(a)

We get:

log2(lgx + 2√lgx + 1) - 2 * (log2(√lgx + 1) / log2(4)) = 1

Step 2: Simplify the expression log2(4).

log2(4) = log2(2^2) = 2 * log2(2) = 2 * 1 = 2

The equation now becomes:

log2(lgx + 2√lgx + 1) - 2 * (log2(√lgx + 1) / 2) = 1

Step 3: Simplify the term log2(√lgx + 1) / 2.

Using the property log_b(a^c) = c * log_b(a), we have:

log2(√lgx + 1) / 2 = (1/2) * log2(√lgx + 1)

The equation becomes:

log2(lgx + 2√lgx + 1) - (1/2) * log2(√lgx + 1) = 1

Step 4: Combine the logarithms.

Using the property log_b(a) - log_b(c) = log_b(a/c), we have:

log2(lgx + 2√lgx + 1) - log2(√lgx + 1)^(1/2) = 1

Step 5: Convert the equation into exponential form.

Recall that log_b(a) = c is equivalent to b^c = a.

For the first term, we have:

lgx + 2√lgx + 1 = 2^(1)

And for the second term, we have:

√lgx + 1 = 2^(1/2)

Step 6: Solve for √lgx.

Let's solve for √lgx + 1 first:

√lgx + 1 = 2^(1/2)

Solving for √lgx:

√lgx = 2^(1/2) - 1

Now, square both sides to get rid of the square root:

(√lgx)^2 = (2^(1/2) - 1)^2

lgx = (2^(1/2) - 1)^2

lgx = 2 - 2 * 2^(1/2) + 1

lgx = 3 - 2 * 2^(1/2)

Step 7: Solve for lgx.

Now, we have:

lgx + 2√lgx + 1 = 2^(1)

Substitute the value of √lgx we obtained:

(3 - 2 * 2^(1/2)) + 2 * (2^(1/2) - 1) + 1 = 2

Combine like terms:

3 - 2 * 2^(1/2) + 2 * 2^(1/2) - 2 + 1 = 2

3 - 1 = 2

2 = 2

The equation is true for all values of x. Therefore, the solution is all real numbers x that satisfy the original equation.