log10x + log10(x+1)=1
In other words
log base 10 + log base 10(x+1)=1
I need to show it algebraically not using calculator
Thanks a lot !
log10x + log10(x+1)=1
log10[x(x+1)] = 1
x(x+1) = 10^1
x^2 + x - 10 = 0
x = (-1 ± √41)/2
but for log10 x, x > 0
so x = (-1 + √41)/2
thanks for you help and especially explanation
To solve the equation log10(x) + log10(x+1) = 1 algebraically, we can use logarithmic properties. Specifically, we can combine the two logarithms into a single logarithm using the product rule.
The equation can be rewritten as log10(x(x+1)) = 1.
Next, we can convert this logarithmic equation into an exponential equation using the definition of logarithms. In base 10, the logarithmic equation log10(a) = b is equivalent to 10^b = a.
Using this, we can rewrite the equation as 10^1 = x(x+1).
Simplifying, we have 10 = x(x+1).
Expanding the right side, we get 10 = x^2 + x.
Now, rearrange the equation to bring everything to one side: x^2 + x - 10 = 0.
This is now a quadratic equation. To solve it, you can either factor the equation or use the quadratic formula.
Factoring: (x + 5)(x - 2) = 0.
Setting each factor equal to zero, x + 5 = 0 or x - 2 = 0.
Solving these equations, we have x = -5 or x = 2.
Therefore, there are two solutions for x: -5 and 2.
You can plug these values back into the initial equation to verify if they satisfy the equation.