Question

3.27^x+1 + 27.3^3x ÷ 3.3^3x+2 - 1/3.27^x+1

Answer 1

Answer:

$3^{3x+4} - 3^{3x+2}+1$

Step-by-step explanation:

Given expression,

$3\times 27^{x+1} +\frac{27\times 3^{3x}}{3\times 3^{3x+2}} - \frac{1}{3}\times 27^{x+1}$

$3\times (3^3)^{x+1} + \frac{3^3\times 3^{3x}}{3\times 3^{3x+2}} - \frac{1}{3}\times (3^3)^{x+1}$

Using $(a^m)^n = a^{mn}$ and $a^m.a^n = a^{m+n}$

$3\times 3^{3x+3} + \frac{3^{3x+3}}{3^{3x+3}}- \frac{1}{3}\times 3^{3x+3}$

$3^{3x+4} + \frac{3^{3x+3}}{3^{3x+3}}- 3^{3x+3-1}$

$3^{3x+4} +1- 3^{3x+2}$

$3^{3x+4} - 3^{3x+2}+1$

Answer 2

The required answer is 6.

Step-by-step explanation:

Consider the provided expression.

$\frac{\left3\cdot 27^{x+1}\:+\:27\cdot 3^{3x}\:\right}{3\cdot \:3^{3x+2}-\:\frac{1}{3}\cdot \:27^{x+1}}$

Now use the exponent rule: $\:a^b\cdot \:a^c=a^{b+c}$

$\frac{\left3\cdot 27^{x}\cdot 27\:+ 27\cdot 3^{3x}\:\right}{3\cdot 9\cdot \:3^{3x}-\:\frac{1}{3}\cdot27\cdot \:27^{x}}$

$\frac{\left81\cdot 27^{x}\:+ 27\cdot 3^{3x}\:\right}{27\cdot \:3^{3x}-\:9\cdot \:27^{x}}$

$\frac{\left81\cdot 3^{3x}\:+ 27\cdot 3^{3x}\:\right}{27\cdot \:3^{3x}-\:9\cdot \:3^{3x}}$

$\frac{3^{3x}(\left81\:+ 27)}{3^{3x}(27-\:9)}$

$\frac{108}{18}$

$6$

Hence, the required answer is 6.

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