Find the value of x which satisfies the relation Log10 3+log10 (4x+1)=log10…

For logarithms with the same base, the product rule logb(m) + logb(n) = logb(mn) combines a sum of logs into the log of a product, and since logb is one-to-one (injective) on positive reals, logb(p) = logb(q) forces p = q. Also, logb(b) = 1 for any valid base b.

1. Given: log10 3 + log10 (4x+1) = log10 (x+1) + 1

2. Rewrite the constant 1 as log10 10, since log10 10 = 1.

3. Apply the product rule to both sides: the left side becomes log10 (3(4x+1)); the right side becomes log10 (10(x+1)).

4. Since log10 is one-to-one, equate the arguments: 3(4x+1) = 10(x+1).

5. Expand both sides: 12x + 3 = 10x + 10.

6. Solve for x: 2x = 7, so x = 7/2.

Cross-check: substitute x = 7/2 back into the original equation. Then 4x+1 = 15 and x+1 = 9/2. The left side gives log10 3 + log10 15 = log10 45. The right side gives log10 (9/2) + 1 = log10 (9/2) + log10 10 = log10 45. Both sides equal log10 45, confirming the result.

Hence x = 7/2.