Question

Log3(x-5)=log3(x+9) log3(5x-3)=3 lg(x^2-x)=1-lg5 log2 x=1+log2 5 log5(x+1) + log5 (x-1) =3log5 2

Answer 1

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Answer 2

To solve these logarithmic equations, we'll handle each one separately:

1. Log3(x-5) = log3(x+9)

Since the bases are the same (logarithms are base 3), you can set the arguments equal to each other:

x - 5 = x + 9

Now, isolate x:

x - x = 9 + 5 0 = 14

This equation has no solution, which means there are no real values of x that satisfy the original equation.

2. Log3(5x-3) = 3

Again, since the bases are the same, set the arguments equal to each other:

5x - 3 = 3

Now, isolate x:

5x = 3 + 3 5x = 6

Divide by 5:

x = 6/5

3. lg(x^2-x) = 1 - lg(5)

First, let's clarify what "lg" means. It's often used to denote the base-10 logarithm, which is more commonly written as "log10." Assuming that, we can rewrite the equation as:

log10(x^2-x) = 1 - log10(5)

Now, let's use properties of logarithms to simplify this equation. Remember that subtracting logarithms is equivalent to dividing their arguments:

log10(x^2-x) = log10(10) - log10(5)

log10(x^2-x) = log10(2)

Now, set the arguments equal to each other:

x^2 - x = 2

Rearrange and solve for x:

x^2 - x - 2 = 0

Factor the quadratic:

(x - 2)(x + 1) = 0

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

x - 2 = 0 --> x = 2 x + 1 = 0 --> x = -1

So, there are two solutions: x = 2 and x = -1.

4. log2(x) = 1 + log2(5)

First, subtract log2(5) from both sides:

log2(x) - log2(5) = 1

Now, use the properties of logarithms to combine the logarithms on the left side:

log2(x/5) = 1

Now, rewrite this equation in exponential form:

x/5 = 2^1

x/5 = 2

Now, multiply both sides by 5 to solve for x:

x = 5 * 2

x = 10

5. log5(x+1) + log5(x-1) = 3 * log5(2)

Apply the logarithmic properties:

log5((x+1)(x-1)) = log5(2^3)

Now, simplify the right side:

log5((x+1)(x-1)) = log5(8)

Set the arguments equal to each other:

(x+1)(x-1) = 8

Now, expand the left side:

x^2 - 1 = 8

Add 1 to both sides:

x^2 = 9

Take the square root of both sides (remembering both positive and negative roots):

x = ±3

So, there are two solutions: x = 3 and x = -3.

To summarize, the solutions to the given equations are as follows:

1. No real solutions.
2. x = 6/5
3. x = 2 and x = -1
4. x = 10
5. x = 3 and x = -3