Question

solve for x. pls be correct i will mark the brainliest who says the ans first

Answer 1

$\large\text{ x=\dfrac{ 6 \log 2-2 \log 3 }{ 3 \log 2+ \log 3 } . }$

$\large \underline{ \underline{ \text{Type of the question} } }$

Complexed exponential equation

Be careful not to make any mistakes in the long calculations.

$\large \underline{ \underline{ \text{Steps required} } }$

$\begin{aligned} &\rightarrow \boxed{ \text{Properties of exponents} } \\\\&\rightarrow \boxed{ \text{Properties of logarithms} }\\\\&\rightarrow \boxed{ \text{Simplification} }\\\\&\rightarrow \boxed{ \text{Properties of logarithms} } \end{aligned}$

$\large \underline{ \underline{ \text{Explanation} } }$

Let's simplify the power by using of properties of exponents.

$\boxed{ \begin{aligned} &\rightarrow \sqrt[4]{ 24^{x+2} }=8\\\\&\rightarrow ( 24^{x+2} )^{ \frac{1}{4} } =8\ [ \because \sqrt[n]{ a^{m} }=a^{ \frac{ m }{ n } }]\\\\&\rightarrow 24^{ x+2 } =8^{4} . \ [ \because \text{Exponent of 4 on both sides.}] \\\end{aligned} }$

Let's apply logarithm on both sides.

$\boxed{ \begin{aligned} &\rightarrow 24^{x+2} =8^{4} \\\\&\rightarrow(x+2)\log 24 = \log 8^{4}\ [\because \log a^{m} = m \log a]\\\\&\rightarrow x+2=\dfrac{\log 8^{4} }{\log 24} . \end{aligned} }$

Now, let's simplify the equation.

$\boxed{\begin{aligned}&\rightarrow x+2= \dfrac{ \log 8^{4} }{\log 24}\\\\&\rightarrow x= \dfrac{ \log 8^{4} -2\log 24}{ \log 24 } . \end{aligned}}$

By properties of the logarithm,

$\boxed{\begin{aligned}&\rightarrow x=\dfrac{4 \log 8-2 \log 24}{ \log 24} \ [\because \log a^{m} =m \log a]\\\\&\rightarrow x=\dfrac{ 4\log 2^{3} -2(\log 2^{3} + \log 3) }{ \log 2^{3} + \log 3 } \ [ \because \log ab= \log a+ \log b]\\\\&\rightarrow x=\dfrac{ 4\cdot 3\log 2-2(3\log 2+ \log 3) }{ 3 \log 2+ \log 3 } \ [ \because \log a^{m} = m \log a]\\\\&\rightarrow x=\dfrac{ 12 \log 2 -6 \log 2 -2 \log 3 }{ 3 \log 2+ \log 3 }\\\\&\rightarrow x=\dfrac{ 6 \log 2 - 2 \log 3 }{ 3 \log 2+ \log 3 } \end{aligned}}$

Hence,

$\cdots \longrightarrow \boxed{ x=\dfrac{ 6 \log 2 - 2 \log 3 }{ 3 \log 2+ \log 3 } .}$

Answer 2

$\rm \bold{ : \to \: \sqrt[4]{24 ^{x + 2} = 8} } \\ \\ \rm \bold{ : \to \: (24 ^{x + 2}) ^{ \frac{1}{4} } = 8 } \\ \\ \boxed{ \sf \bold{ \gray{24 ^{x + 2} = {8}^{4} }}}$

Using logarithm on both sides.

= 24^x+2 = 8⁴

= (x+2) log 24 = log 8⁴

= x+2 = log8⁴/ log 24

Now simplify the equation

= x+2 = log8⁴/log 24

= x = log8⁴-2log 24/ log 24

Logarithm properties.

= x = 4 log 8 -.2 log 24/log 24

= x = 4 log2³ - 2 (log2³ + log 3)/3 log 2t log

= x = 4.3 log 2 - 6 log 2-2 log2³/3log 2 + log 3

= x = 12 log 2-6 log 2-2 log 3/3log 2 + log 3

= x = 6log 2-2 log 3/3 log 2 + log3