Question

the solution to the equation $6(2^{x + 4}) = 36$ is
(1) $-1$
(2) $\frac{ln 36}{ln 12} - 4$
(3) $ln(3) - 4$
(4) $\frac{ln 6}{ln 2} - 4$
20.
which expression is equivalent to $(x - 2)^2 + 27(x - 2) - 90$?
(1) $(x + 30)(x - 3)$
(2) $(x + 28)(x - 5)$
(3) $(x - 30)(x + 3)$
(4) $(x - 2)(x + 25)(x - 90)$

Explanation:

First Problem:

Step1: Isolate the exponential term

Divide both sides by 6:
$$2^{x+4} = \frac{36}{6} = 6$$

Step2: Take natural log of both sides

Use logarithm power rule $\ln(a^b)=b\ln(a)$:
$$\ln(2^{x+4}) = \ln(6) \implies (x+4)\ln(2) = \ln(6)$$

Step3: Solve for $x+4$

Divide both sides by $\ln(2)$:
$$x+4 = \frac{\ln(6)}{\ln(2)}$$

Step4: Isolate $x$

Subtract 4 from both sides:
$$x = \frac{\ln(6)}{\ln(2)} - 4$$

Step1: Substitute $u = x-2$

Rewrite the expression in terms of $u$:
$$u^2 + 27u - 90$$

Step2: Factor the quadratic

Find two numbers that multiply to $-90$ and add to $27$ (30 and $-3$):
$$u^2 + 27u - 90 = (u+30)(u-3)$$

Step3: Substitute back $u=x-2$

Replace $u$ with the original expression:
$$(x-2+30)(x-2-3) = (x+28)(x-5)$$

Answer:

(4) $\frac{\ln 6}{\ln 2} - 4$

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Second Problem: