How do you solve #Log(x+2)+log(x-1)=1#?

Question

Lincoln Anderson · Accepted Answer

#x=-4#

The basic approach to solving log equations it to get them into simplest form and then put them back into exponential form.

In the case of this problem, the log expressions are not as simple as possible and you need to use your log rules to combine the two expressions. Because there is not base written with LOG the base is assumed to be 10.

#log[(x+2)*(x-1)]=1# #log[x^2+x-2]=1#

put the equation back into exponential form:

#10^1=x^2+x-2# #10=x^2+x-2# #x^2+x-12=0# #(x+4)(x-3)=0# #x=-4 and 3#

Issues of domain exist with log equation since the domain of #y=logx# is x>0, therefore it is crucial to test answers to ensure they work. In this case, #x=-4# makes each log expression undefined therefore the only answer is #x=3#.

Ariana Alvarez · Answer

Assuming the base of the logarithm is 10, the answer is 3.

Using the fact that log X + log Y = log XY, one gets #log(x+2)+log(x-1)=log((x+2)(x-1)) = log(x^2+x-2)=1#. That means #10^(log(x^2+x-2)) = 10^1#. So, by the definition of a logarithm, one gets #x^2+x-2 = 10#, or #x^2+x-12 = 0#. By the quadratic equation, one gets #x =( (-1 (+-) sqrt(1-4*1*(-12))) /2 )#. That is, x = #(-1 (+/-) sqrt (49))/2# or x = -4 or x = 3. Since x can't be -4, as the logarithm of a negative number is undefined, one gets x=3.

Serenity Jones · Answer

undefined To solve the equation $ \log(x+2) + \log(x-1) = 1 $:

1. Combine the logarithms using the product rule of logarithms, which states that $ \log(a) + \log(b) = \log(ab) $:
$$ \log((x+2)(x-1)) = 1 $$

2. Rewrite the equation in exponential form. Remember that if $ \log_b(x) = y $, then $ b^y = x $:
$$ 10^1 = (x+2)(x-1) $$

3. Simplify the equation:
$$ 10 = (x+2)(x-1) $$

4. Expand and rearrange the equation:
$$ 10 = x^2 + x - 2 $$
$$ x^2 + x - 12 = 0 $$

5. Solve the quadratic equation. You can factorize or use the quadratic formula:
$$ x^2 + 4x - 3x - 12 = 0 $$
$$ x(x + 4) - 3(x + 4) = 0 $$
$$ (x - 3)(x + 4) = 0 $$

This gives two solutions:
$$ x - 3 = 0 \rightarrow x = 3 $$
$$ x + 4 = 0 \rightarrow x = -4 $$

However, we need to check if these solutions satisfy the original equation. Since logarithms are only defined for positive numbers, we discard the solution $ x = -4 $.

Thus, the only solution to the equation $ \log(x+2) + \log(x-1) = 1 $ is $ x = 3 $.