Question

5-35.
solve each of the following equations for ( x ). give exact answers using fractions, not decimals.

a. ( 8^{(x+3)} = 32 )

step 1(a):

b. ( 27^{2x} = left( \frac{1}{9}
ight)^{(x-1)} )

hint (b):

c. ( log_{3}(x) = 4 )

hint (c):

d. ( log_{x}(64) = 3 )

step (d):

Explanation:

Part a:

Step1: Rewrite bases as powers of 2

$8 = 2^3$, $32 = 2^5$, so:
$$(2^3)^{(x+3)} = 2^5$$

Step2: Simplify exponent rule

$(a^m)^n = a^{mn}$, so:
$$2^{3(x+3)} = 2^5$$

Step3: Set exponents equal

If $a^m=a^n$ ($a>0,a
eq1$), $m=n$:
$$3(x+3) = 5$$

Step4: Solve for x

Expand and isolate x:
$$3x + 9 = 5 \implies 3x = 5-9 \implies 3x=-4 \implies x=-\frac{4}{3}$$

Part b:

Step1: Rewrite bases as powers of 3

$27=3^3$, $\frac{1}{9}=3^{-2}$, so:
$$(3^3)^{2x} = (3^{-2})^{(x-1)}$$

Step2: Simplify exponent rule

$(a^m)^n = a^{mn}$, so:
$$3^{6x} = 3^{-2(x-1)}$$

Step3: Set exponents equal

If $a^m=a^n$ ($a>0,a
eq1$), $m=n$:
$$6x = -2(x-1)$$

Step4: Solve for x

Expand and isolate x:
$$6x = -2x + 2 \implies 6x+2x=2 \implies 8x=2 \implies x=\frac{2}{8}=\frac{1}{4}$$

Part c:

Step1: Convert log to exponential form

$\log_b(a)=c \implies b^c=a$, so:
$$x = 3^4$$

Step2: Calculate the value

$$x=81$$

Part d:

Step1: Convert log to exponential form

$\log_x(64)=3 \implies x^3=64$

Step2: Solve for x

Find cube root of 64:
$$x=\sqrt[3]{64}=4$$

Answer:

a. $x=-\frac{4}{3}$
b. $x=\frac{1}{4}$
c. $x=81$
d. $x=4$