Question

Lg(x+1)+lg(x+4)=1 ???

Answer 1

Lg(x+1)+lg(x+4)=1
lg(x+1)+lg(x+4)=lg10
(ОДЗ: х+1>0;х>-1
х+4>0;х>-4)
lg(x+1)(x+4)=lg10
x^2+5x+4=10
x^2+5x-6=0
х=-6 и х=1
х=-6 (не подходит по ОДЗ)
Ответ: х = 1

Answer 2

I hope this helps you

Answer 3

To solve the equation lg(x+1) + lg(x+4) = 1, first, we need to clarify that "lg" refers to the logarithm with base 10. The equation can be rewritten using logarithm properties:

lg(x+1) + lg(x+4) = 1 lg((x+1)(x+4)) = 1

Now, we can remove the logarithm by converting the equation back to exponential form:

10^1 = (x+1)(x+4) 10 = (x+1)(x+4)

Next, we can expand the right side:

10 = x^2 + 5x + 4

Now, we have a quadratic equation, which can be rearranged to the standard form:

x^2 + 5x + 4 - 10 = 0 x^2 + 5x - 6 = 0

To solve for x, we can factor the quadratic equation:

(x + 6)(x - 1) = 0

Now, set each factor to zero and solve for x:

1. x + 6 = 0 x = -6

2. x - 1 = 0 x = 1

So, the solutions to the equation are x = -6 and x = 1. However, since the logarithm requires a positive argument, the solution x = -6 is extraneous, and the valid solution is x = 1.