Solve the following logarithmic equations: 1) log₂(x) + log₂(x−2) = 1 2) log₅(x+1) - log₅(x−1) = 1 3) log₃(x)…

Question

Solve the following logarithmic equations: 1) log₂(x) + log₂(x−2) = 1 2) log₅(x+1) - log₅(x−1) = 1 3) log₃(x) + 2·log₃(x) = 6 4) log₄(x²−4) - log₄(x+2) = 1 5) log₂(x) + log₄(x) = 3 6) log(x²−1) = 2·log(x+1) 7) log₃(x) + log₃(x−2) + log₃(x−4) = log₃(15)

ScanSolve AI Tutor · Accepted Answer

Here are the solutions to the logarithm equations:

1) $\log_2 x + \log_2 (x-2) = 1$

Use the logarithm property $\log_b M + \log_b N = \log_b (MN)$ .
$\log_2 (x(x-2)) = 1$
Convert the logarithmic equation to an exponential equation ( $M = b^N$ ).
$x(x-2) = 2^1$ $x^2 - 2x = 2$ $x^2 - 2x - 2 = 0$
Solve the quadratic equation using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ .
$x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-2)}}{2(1)}$ $x = \frac{2 \pm \sqrt{4 + 8}}{2}$ $x = \frac{2 \pm \sqrt{12}}{2}$ $x = \frac{2 \pm 2\sqrt{3}}{2}$ $x = 1 \pm \sqrt{3}$
Check the domain restriction ( $x > 0$ and $x-2 > 0 \implies x > 2$ ).
$x = 1 + \sqrt{3} \approx 2.732$ is valid.
$x = 1 - \sqrt{3} \approx -0.732$ is not valid.
The solution is $\boxed{x = 1 + \sqrt{3}}$ .

2) $\log_5 (x+1) - \log_5 (x-1) = 1$
Use the logarithm property $\log_b M - \log_b N = \log_b \left(\frac{M}{N}\right)$ .
$\log_5 \left(\frac{x+1}{x-1}\right) = 1$
Convert to an exponential equation.
$\frac{x+1}{x-1} = 5^1$ $x+1 = 5(x-1)$ $x+1 = 5x - 5$
Solve for $x$ .
$6 = 4x$ $x = \frac{6}{4}$ $x = \frac{3}{2}$
Check the domain restriction ( $x+1 > 0 \implies x > -1$ and $x-1 > 0 \implies x > 1$ , so $x > 1$ ).
$x = \frac{3}{2} = 1.5$ is valid.
The solution is $\boxed{x = \frac{3}{2}}$ .

3) $\log_3 x + 2 \cdot \log_3 x = 6$
Combine like terms.
$3 \cdot \log_3 x = 6$
Isolate $\log_3 x$ .
$\log_3 x = \frac{6}{3}$ $\log_3 x = 2$
Convert to an exponential equation.
$x = 3^2$ $x = 9$
Check the domain restriction ( $x > 0$ ).
$x = 9$ is valid.
The solution is $\boxed{x = 9}$ .

4) $\log_4 (x^2-4) - \log_4 (x+2) = 1$
Use the logarithm property $\log_b M - \log_b N = \log_b \left(\frac{M}{N}\right)$ .
$\log_4 \left(\frac{x^2-4}{x+2}\right) = 1$
Factor the numerator $x^2-4 = (x-2)(x+2)$ .
$\log_4 \left(\frac{(x-2)(x+2)}{x+2}\right) = 1$
Simplify the expression inside the logarithm.
The domain requires $x > 2$ , so $x+2 \neq 0$ .
$\log_4 (x-2) = 1$
Convert to an exponential equation.
$x-2 = 4^1$ $x-2 = 4$ $x = 6$
Check the domain restriction ( $x^2-4 > 0 \implies x < -2 \text{ or } x > 2$ , and $x+2 > 0 \implies x > -2$ .
Combined, $x > 2$ ).
$x = 6$ is valid.
The solution is $\boxed{x = 6}$ .

5) $\log_2 x + \log_4 x = 3$
Change the base of $\log_4 x$ to base 2 using $\log_b M = \frac{\log_c M}{\log_c b}$ .
$\log_4 x = \frac{\log_2 x}{\log_2 4} = \frac{\log_2 x}{2}$
Substitute this into the original equation.
$\log_2 x + \frac{\log_2 x}{2} = 3$
Combine the terms with $\log_2 x$ .
$\left(1 + \frac{1}{2}\right) \log_2 x = 3$ $\frac{3}{2} \log_2 x = 3$
Isolate $\log_2 x$ .
$\log_2 x = 3 \cdot \frac{2}{3}$ $\log_2 x = 2$
Convert to an exponential equation.
$x = 2^2$ $x = 4$
Check the domain restriction ( $x > 0$ ).
$x = 4$ is valid.
The solution is $\boxed{x = 4}$ .

6) $\log (x^2-1) = 2 \cdot \log (x+1)$
Use the logarithm property $k \log_b M = \log_b (M^k)$ on the right side.
$\log (x^2-1) = \log ((x+1)^2)$
If $\log_b M = \log_b N$ , then $M = N$ .
$x^2-1 = (x+1)^2$
Expand the right side and solve for $x$ .
$x^2-1 = x^2 + 2x + 1$ $-1 = 2x + 1$ $-2 = 2x$ $x = -1$
Check the domain restriction ( $x^2-1 > 0 \implies x < -1 \text{ or } x > 1$ , and $x+1 > 0 \implies x > -1$ .
Combined, $x > 1$ ).
$x = -1$ is not greater than 1. Also, $\log(x+1)$ would be $\log(0)$ , which is undefined.
The equation has $\boxed{\text{no solution}}$ .

7) $\log_3 x + \log_3 (x-2) + \log_3 (x-4) = \log_3 (15)$
Use the logarithm property $\log_b M + \log_b N = \log_b (MN)$ on the left side.
$\log_3 (x(x-2)(x-4)) = \log_3 (15)$
If $\log_b M = \log_b N$ , then $M = N$ .
$x(x-2)(x-4) = 15$
Expand the left side.
$x(x^2 - 4x - 2x + 8) = 15$ $x(x^2 - 6x + 8) = 15$ $x^3 - 6x^2 + 8x = 15$ $x^3 - 6x^2 + 8x - 15 = 0$
Find integer roots using the Rational Root Theorem.
Test divisors of 15.
Let $P(x) = x^3 - 6x^2 + 8x - 15$ .
$P(5) = 5^3 - 6(5^2) + 8(5) - 15 = 125 - 150 + 40 - 15 = 0$ . So, $x=5$ is a root.
Perform polynomial division to factor $(x-5)$ out.
$(x-5)(x^2 - x + 3) = 0$
Solve the quadratic factor $x^2 - x + 3 = 0$ .
The discriminant is $\Delta = (-1)^2 - 4(1)(3) = 1 - 12 = -11$ . Since $\Delta < 0$ , there are no other real roots.
Check the domain restriction ( $x > 0$ , $x-2 > 0 \implies x > 2$ , and $x-4 > 0 \implies x > 4$ .
Combined, $x > 4$ ).
$x = 5$ is valid.
The solution is $\boxed{x = 5}$ .

Solve the following logarithmic equations: 1) log₂(x) + log₂(x−2) = 1 2) log₅(x+1) − log₅(x−1) = 1 3) log₃(x) + 2·log₃(x) = 6 4) log₄(x²−4) − log₄(x+2) = 1 5) log₂(x) + log₄(x) = 3 6) log(x²−1) = 2·log(x+1) 7) log₃(x) + log₃(x-2) + log₃(x-4) = log₃(15)

Step-by-step solution

Use the logarithm property $\log_b M + \log_b N = \log_b (MN)$ .

Convert the logarithmic equation to an exponential equation ( $M = b^N$ ).

Solve the quadratic equation using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ .

Check the domain restriction ( $x > 0$ and $x-2 > 0 \implies x > 2$ ).

Use the logarithm property $\log_b M - \log_b N = \log_b \left(\frac{M}{N}\right)$ .

Convert to an exponential equation.

Solve for $x$ .

Check the domain restriction ( $x+1 > 0 \implies x > -1$ and $x-1 > 0 \implies x > 1$ , so $x > 1$ ).

Combine like terms.

Isolate $\log_3 x$ .

Convert to an exponential equation.

Check the domain restriction ( $x > 0$ ).

Use the logarithm property $\log_b M - \log_b N = \log_b \left(\frac{M}{N}\right)$ .

Factor the numerator $x^2-4 = (x-2)(x+2)$ .

Simplify the expression inside the logarithm.

Convert to an exponential equation.

Check the domain restriction ( $x^2-4 > 0 \implies x < -2 \text{ or } x > 2$ , and $x+2 > 0 \implies x > -2$ .

Change the base of $\log_4 x$ to base 2 using $\log_b M = \frac{\log_c M}{\log_c b}$ .

Substitute this into the original equation.

Combine the terms with $\log_2 x$ .

Isolate $\log_2 x$ .

Convert to an exponential equation.

Check the domain restriction ( $x > 0$ ).

Use the logarithm property $k \log_b M = \log_b (M^k)$ on the right side.

If $\log_b M = \log_b N$ , then $M = N$ .

Expand the right side and solve for $x$ .

Check the domain restriction ( $x^2-1 > 0 \implies x < -1 \text{ or } x > 1$ , and $x+1 > 0 \implies x > -1$ .

Use the logarithm property $\log_b M + \log_b N = \log_b (MN)$ on the left side.

If $\log_b M = \log_b N$ , then $M = N$ .

Expand the left side.

Find integer roots using the Rational Root Theorem.

Perform polynomial division to factor $(x-5)$ out.

Solve the quadratic factor $x^2 - x + 3 = 0$ .

Check the domain restriction ( $x > 0$ , $x-2 > 0 \implies x > 2$ , and $x-4 > 0 \implies x > 4$ .

Still stuck on this one?

More Mathematics Questions

Solve the following logarithmic equations: 1) log₂(x) + log₂(x−2) = 1 2) log₅(x+1) − log₅(x−1) = 1 3) log₃(x) + 2·log₃(x) = 6 4) log₄(x²−4) − log₄(x+2) = 1 5) log₂(x) + log₄(x) = 3 6) log(x²−1) = 2·log(x+1) 7) log₃(x) + log₃(x-2) + log₃(x-4) = log₃(15)

Step-by-step solution

Use the logarithm property log⁡bM+log⁡bN=log⁡b(MN)\log_b M + \log_b N = \log_b (MN)logb​M+logb​N=logb​(MN).

Convert the logarithmic equation to an exponential equation (M=bNM = b^NM=bN).

Solve the quadratic equation using the quadratic formula x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}x=2a−b±b2−4ac​​.

Check the domain restriction (x>0x > 0x>0 and x−2>0 ⟹ x>2x-2 > 0 \implies x > 2x−2>0⟹x>2).

Use the logarithm property log⁡bM−log⁡bN=log⁡b(MN)\log_b M - \log_b N = \log_b \left(\frac{M}{N}\right)logb​M−logb​N=logb​(NM​).

Convert to an exponential equation.

Solve for xxx.

Check the domain restriction (x+1>0 ⟹ x>−1x+1 > 0 \implies x > -1x+1>0⟹x>−1 and x−1>0 ⟹ x>1x-1 > 0 \implies x > 1x−1>0⟹x>1, so x>1x > 1x>1).

Combine like terms.

Isolate log⁡3x\log_3 xlog3​x.

Convert to an exponential equation.

Check the domain restriction (x>0x > 0x>0).

Use the logarithm property log⁡bM−log⁡bN=log⁡b(MN)\log_b M - \log_b N = \log_b \left(\frac{M}{N}\right)logb​M−logb​N=logb​(NM​).

Factor the numerator x2−4=(x−2)(x+2)x^2-4 = (x-2)(x+2)x2−4=(x−2)(x+2).

Simplify the expression inside the logarithm.

Convert to an exponential equation.

Check the domain restriction (x2−4>0 ⟹ x<−2 or x>2x^2-4 > 0 \implies x < -2 \text{ or } x > 2x2−4>0⟹x<−2 or x>2, and x+2>0 ⟹ x>−2x+2 > 0 \implies x > -2x+2>0⟹x>−2.

Change the base of log⁡4x\log_4 xlog4​x to base 2 using log⁡bM=log⁡cMlog⁡cb\log_b M = \frac{\log_c M}{\log_c b}logb​M=logc​blogc​M​.

Substitute this into the original equation.

Combine the terms with log⁡2x\log_2 xlog2​x.

Isolate log⁡2x\log_2 xlog2​x.

Convert to an exponential equation.

Check the domain restriction (x>0x > 0x>0).

Use the logarithm property klog⁡bM=log⁡b(Mk)k \log_b M = \log_b (M^k)klogb​M=logb​(Mk) on the right side.

If log⁡bM=log⁡bN\log_b M = \log_b Nlogb​M=logb​N, then M=NM = NM=N.

Expand the right side and solve for xxx.

Check the domain restriction (x2−1>0 ⟹ x<−1 or x>1x^2-1 > 0 \implies x < -1 \text{ or } x > 1x2−1>0⟹x<−1 or x>1, and x+1>0 ⟹ x>−1x+1 > 0 \implies x > -1x+1>0⟹x>−1.

Use the logarithm property log⁡bM+log⁡bN=log⁡b(MN)\log_b M + \log_b N = \log_b (MN)logb​M+logb​N=logb​(MN) on the left side.

If log⁡bM=log⁡bN\log_b M = \log_b Nlogb​M=logb​N, then M=NM = NM=N.

Expand the left side.

Find integer roots using the Rational Root Theorem.

Perform polynomial division to factor (x−5)(x-5)(x−5) out.

Solve the quadratic factor x2−x+3=0x^2 - x + 3 = 0x2−x+3=0.

Check the domain restriction (x>0x > 0x>0, x−2>0 ⟹ x>2x-2 > 0 \implies x > 2x−2>0⟹x>2, and x−4>0 ⟹ x>4x-4 > 0 \implies x > 4x−4>0⟹x>4.

Still stuck on this one?

More Mathematics Questions

Use the logarithm property $\log_b M + \log_b N = \log_b (MN)$ .

Convert the logarithmic equation to an exponential equation ( $M = b^N$ ).

Solve the quadratic equation using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ .

Check the domain restriction ( $x > 0$ and $x-2 > 0 \implies x > 2$ ).

Use the logarithm property $\log_b M - \log_b N = \log_b \left(\frac{M}{N}\right)$ .

Solve for $x$ .

Check the domain restriction ( $x+1 > 0 \implies x > -1$ and $x-1 > 0 \implies x > 1$ , so $x > 1$ ).

Isolate $\log_3 x$ .

Check the domain restriction ( $x > 0$ ).

Use the logarithm property $\log_b M - \log_b N = \log_b \left(\frac{M}{N}\right)$ .

Factor the numerator $x^2-4 = (x-2)(x+2)$ .

Check the domain restriction ( $x^2-4 > 0 \implies x < -2 \text{ or } x > 2$ , and $x+2 > 0 \implies x > -2$ .

Change the base of $\log_4 x$ to base 2 using $\log_b M = \frac{\log_c M}{\log_c b}$ .

Combine the terms with $\log_2 x$ .

Isolate $\log_2 x$ .

Check the domain restriction ( $x > 0$ ).

Use the logarithm property $k \log_b M = \log_b (M^k)$ on the right side.

If $\log_b M = \log_b N$ , then $M = N$ .

Expand the right side and solve for $x$ .

Check the domain restriction ( $x^2-1 > 0 \implies x < -1 \text{ or } x > 1$ , and $x+1 > 0 \implies x > -1$ .

Use the logarithm property $\log_b M + \log_b N = \log_b (MN)$ on the left side.

If $\log_b M = \log_b N$ , then $M = N$ .

Perform polynomial division to factor $(x-5)$ out.

Solve the quadratic factor $x^2 - x + 3 = 0$ .

Check the domain restriction ( $x > 0$ , $x-2 > 0 \implies x > 2$ , and $x-4 > 0 \implies x > 4$ .