Question

question 1 of 5
solve for $x$.
$9^{-x-5}=27$

question 2 of 5
solve for $x$.
$64=16^{x+1}$

question 3 of 5
solve for $x$.
$3^{12x}=27^{3x+1}$

Explanation:

Question 1:

Step1: Rewrite bases as powers of 3

$9=3^2$, $27=3^3$, so:
$$(3^2)^{-x-5}=3^3$$

Step2: Simplify left side exponent

Use $(a^m)^n=a^{m \cdot n}$:
$$3^{2(-x-5)}=3^3$$
$$3^{-2x-10}=3^3$$

Step3: Set exponents equal

Since bases are equal, exponents are equal:
$$-2x-10=3$$

Step4: Solve for x

Add 10 to both sides:
$$-2x=13$$
Divide by -2:
$$x=-\frac{13}{2}$$

Question 2:

Step1: Rewrite bases as powers of 4

$64=4^3$, $16=4^2$, so:
$$4^3=(4^2)^{x+1}$$

Step2: Simplify right side exponent

Use $(a^m)^n=a^{m \cdot n}$:
$$4^3=4^{2(x+1)}$$
$$4^3=4^{2x+2}$$

Step3: Set exponents equal

Since bases are equal, exponents are equal:
$$3=2x+2$$

Step4: Solve for x

Subtract 2 from both sides:
$$1=2x$$
Divide by 2:
$$x=\frac{1}{2}$$

Question 3:

Step1: Rewrite 27 as power of 3

$27=3^3$, so:
$$3^{12x}=(3^3)^{3x+1}$$

Step2: Simplify right side exponent

Use $(a^m)^n=a^{m \cdot n}$:
$$3^{12x}=3^{3(3x+1)}$$
$$3^{12x}=3^{9x+3}$$

Step3: Set exponents equal

Since bases are equal, exponents are equal:
$$12x=9x+3$$

Step4: Solve for x

Subtract $9x$ from both sides:
$$3x=3$$
Divide by 3:
$$x=1$$

Answer:

Question 1: $x=-\frac{13}{2}$
Question 2: $x=\frac{1}{2}$
Question 3: $x=1$