Solve for m log base 4 of 2m^3-14m^2- log base 4 of 2m = log base 4 of 8

$\log_{4} (2 m^{3} - 14 m^{2}) - \log_{4} (2 m) = \log_{4} (8)$

Simplify the left side.

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$\log_{4} (\frac{2 m^{3} - 14 m^{2}}{2 m}) = \log_{4} (8)$

Factor $2 m^{2}$ out of $2 m^{3} - 14 m^{2}$ .

Factor $2 m^{2}$ out of $2 m^{3}$ .

$\log_{4} (\frac{2 m^{2} (m) - 14 m^{2}}{2 m}) = \log_{4} (8)$

Factor $2 m^{2}$ out of $- 14 m^{2}$ .

$\log_{4} (\frac{2 m^{2} (m) + 2 m^{2} (- 7)}{2 m}) = \log_{4} (8)$

Factor $2 m^{2}$ out of $2 m^{2} (m) + 2 m^{2} (- 7)$ .

$\log_{4} (\frac{2 m^{2} (m - 7)}{2 m}) = \log_{4} (8)$

Cancel the common factor of $2$ .

Cancel the common factor.

$\log_{4} (\frac{2 m^{2} (m - 7)}{2 m}) = \log_{4} (8)$

Rewrite the expression.

$\log_{4} (\frac{m^{2} (m - 7)}{m}) = \log_{4} (8)$

Cancel the common factor of $m^{2}$ and $m$ .

Factor $m$ out of $m^{2} (m - 7)$ .

$\log_{4} (\frac{m (m (m - 7))}{m}) = \log_{4} (8)$

Cancel the common factors.

Raise $m$ to the power of $1$ .

$\log_{4} (\frac{m (m (m - 7))}{m}) = \log_{4} (8)$

Factor $m$ out of $m^{1}$ .

$\log_{4} (\frac{m (m (m - 7))}{m \cdot 1}) = \log_{4} (8)$

Cancel the common factor.

$\log_{4} (\frac{m (m (m - 7))}{m \cdot 1}) = \log_{4} (8)$

Rewrite the expression.

$\log_{4} (\frac{m (m - 7)}{1}) = \log_{4} (8)$

Divide $m (m - 7)$ by $1$ .

$\log_{4} (m (m - 7)) = \log_{4} (8)$

Apply the distributive property.

$\log_{4} (m \cdot m + m \cdot - 7) = \log_{4} (8)$

Simplify the expression.

Multiply $m$ by $m$ .

$\log_{4} (m^{2} + m \cdot - 7) = \log_{4} (8)$

Move $- 7$ to the left of $m$ .

$\log_{4} (m^{2} - 7 m) = \log_{4} (8)$

Logarithm base $4$ of $8$ is $\frac{3}{2}$ .

Rewrite as an equation.

$\log_{4} (8) = x$

Rewrite $\log_{4} (8) = x$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b$ does not equal $1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

$4^{x} = 8$

Create expressions in the equation that all have equal bases.

${(2^{2})}^{x} = 2^{3}$

Rewrite ${(2^{2})}^{x}$ as $2^{2 x}$ .

$2^{2 x} = 2^{3}$

Since the bases are the same, then two expressions are only equal if the exponents are also equal.

$2 x = 3$

Solve for $x$ .

$x = \frac{3}{2}$

The variable $x$ is equal to $\frac{3}{2}$ .

$\log_{4} (m^{2} - 7 m) = \frac{3}{2}$

Rewrite $\log_{4} (m^{2} - 7 m) = \frac{3}{2}$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b \neq 1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

$4^{\frac{3}{2}} = m^{2} - 7 m$

Solve for $m$ .

Rewrite the equation as $m^{2} - 7 m = 4^{\frac{3}{2}}$ .

$m^{2} - 7 m = 4^{\frac{3}{2}}$

Simplify $4^{\frac{3}{2}}$ .

Simplify the expression.

Rewrite $4$ as $2^{2}$ .

$m^{2} - 7 m = {(2^{2})}^{\frac{3}{2}}$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$m^{2} - 7 m = 2^{2 (\frac{3}{2})}$

Cancel the common factor of $2$ .

Cancel the common factor.

$m^{2} - 7 m = 2^{2 (\frac{3}{2})}$

Rewrite the expression.

$m^{2} - 7 m = 2^{3}$

Raise $2$ to the power of $3$ .

$m^{2} - 7 m = 8$

Subtract $8$ from both sides of the equation.

$m^{2} - 7 m - 8 = 0$

Factor $m^{2} - 7 m - 8$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 8$ and whose sum is $- 7$ .

$- 8, 1$

Write the factored form using these integers.

$(m - 8) (m + 1) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$m - 8 = 0$

$m + 1 = 0$

Set $m - 8$ equal to $0$ and solve for $m$ .

Set $m - 8$ equal to $0$ .

$m - 8 = 0$

Add $8$ to both sides of the equation.

$m = 8$

Set $m + 1$ equal to $0$ and solve for $m$ .

Set $m + 1$ equal to $0$ .

$m + 1 = 0$

Subtract $1$ from both sides of the equation.

$m = - 1$

The final solution is all the values that make $(m - 8) (m + 1) = 0$ true.

$m = 8, - 1$

Exclude the solutions that do not make $\log_{4} (2 m^{3} - 14 m^{2}) - \log_{4} (2 m) = \log_{4} (8)$ true.

$m = 8$

Solve for m log base 4 of 2m^3-14m^2- log base 4 of 2m = log base 4 of 8

Related Questions

Support LCMGCF.com by disabling your adblocker.