Question

find x if root (5^0 + 2/3)=(0.6)^2-3x

Answer 1

SOLUTION

TO DETERMINE

The value of x when

$\displaystyle \sf{ \sqrt{ \bigg( {5}^{0} + \frac{2}{3} \bigg) } = {(0.6)}^{2 - 3x} \: }$

EVALUATION

$\displaystyle \sf{ \sqrt{ \bigg( {5}^{0} + \frac{2}{3} \bigg) } = {(0.6)}^{2 - 3x} \: }$

$\implies \displaystyle \sf{ \sqrt{ \bigg( 1 + \frac{2}{3} \bigg) } = { \bigg( \frac{3}{5} \bigg)}^{2 - 3x} \: }$

$\implies \displaystyle \sf{ \sqrt{ \bigg( \frac{5}{3} \bigg) } = { \bigg( \frac{5}{3} \bigg)}^{3x - 2} \: }$

$\implies \displaystyle \sf{ {{ \bigg( \frac{5}{3} \bigg)}^{ \frac{1}{2} } } = { \bigg( \frac{5}{3} \bigg)}^{3x - 2} \: }$

$\implies \displaystyle \sf{ \frac{1}{2} = 3x - 2}$

$\implies \displaystyle \sf{ 3x = 2 + \frac{1}{2} }$

$\implies \displaystyle \sf{ 3x = \frac{5}{2} }$

$\implies \displaystyle \sf{ x = \frac{5}{6} }$

FINAL ANSWER

$\boxed{ \displaystyle \sf{ \: \: x = \frac{5}{6} } \: \: }$

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If 2^x+3 + 2^x – 4 =129,

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Answer 2

$\underline{\underline{\sf{\clubsuit\:\:Question}}}$

• Find "x" if  $\sf{\sqrt{\left(5^0\:+\:\dfrac{2}{3}\right)}=\left(0.6\right)^{2-3x}}$

$\underline{\underline{\sf{\clubsuit\:\:Answer}}}$

$\bf{\bigstar \:\:\:x=\dfrac{5}{6}}$

$\underline{\underline{\sf{\clubsuit\:\:Calculations}}}$

$\sf{\sqrt{\left(5^0\:+\:\dfrac{2}{3}\right)}=\left(0.6\right)^{2-3x}}$

Apply exponent rule : $\sf{\sqrt{a}=a^{\tfrac{1}{2}}}$

$\sf{\left(5^0+\dfrac{2}{3}\right)^{\tfrac{1}{2}}=0.6^{2-3x}}$

$\sf{\implies\left(5^0+\dfrac{2}{3}\right)^{\frac{1}{2}}=\left(\dfrac{6}{10}\right)^{2-3x}}$

$\sf{\implies\left(5^0+\dfrac{2}{3}\right)^{\frac{1}{2}}=\left(\dfrac{3}{5}\right)^{2-3x}}$

$\sf{\implies \sqrt{5^0+\dfrac{2}{3}}=\left(\dfrac{3}{5}\right)^{2-3x}}$

$\sf{\implies\:\displaystyle\sqrt{5^0+\frac{2}{3}}=\left(\left(5^0+\frac{2}{3}\right)^{-1}\right)^{2-3x}}$

Again Apply exponent rule : $\sf{\sqrt{a}=a^{\tfrac{1}{2}}}$

$\sf{\implies\displaystyle\left(5^0+\frac{2}{3}\right)^{\frac{1}{2}}=\left(\left(5^0+\frac{2}{3}\right)^{-1}\right)^{2-3x}}$

Apply exponent rule : $\sf{\left(a^{b}\right)^{c}=a^{b c}}$

$\sf{\implies\displaystyle\left(5^0+\frac{2}{3}\right)^{\frac{1}{2}}=\left(5^0+\frac{2}{3}\right)^{-1\cdot \left(2-3x\right)}}$

$\Large\boxed{\sf{\mathrm{If\:}a^{f\left(x\right)}=a^{g\left(x\right)}\mathrm{,\:then\:}f\left(x\right)=g\left(x\right)}}$

$\sf{\implies\dfrac{1}{2}=-1\cdot \left(2-3x\right)}$

$\sf{\implies\dfrac{1}{2}=-\left(2-3x\right)}$

Siwtch Sides :

$\sf{\implies-\left(2-3x\right)=\dfrac{1}{2}}$

$\sf{\implies\dfrac{-\left(2-3x\right)}{-1}=\dfrac{\tfrac{1}{2}}{-1}}$

$\sf{\implies\displaystyle2-3x=-\frac{1}{2}}$

$\sf{\implies2-3x-2=-\dfrac{1}{2}-2}$

$\sf{\implies-3x=-\dfrac{5}{2}}$

$\sf{\implies\dfrac{-3x}{-3}=\dfrac{-\tfrac{5}{2}}{-3}}$

$\boxed{\sf{\implies x=\frac{5}{6}}}$