Step-by-step explanation:

Step-by-step explanation:Solution:

Step-by-step explanation:Solution:(0.2)x = 2.

Step-by-step explanation:Solution:(0.2)x = 2.Taking log on both sides

Step-by-step explanation:Solution:(0.2)x = 2.Taking log on both sideslog (0.2)x = log 2.

Step-by-step explanation:Solution:(0.2)x = 2.Taking log on both sideslog (0.2)x = log 2.x log (0.2) = 0.3010, [since log 2 = 0.3010].

Step-by-step explanation:Solution:(0.2)x = 2.Taking log on both sideslog (0.2)x = log 2.x log (0.2) = 0.3010, [since log 2 = 0.3010].x log (2/10) = 0.3010.

Step-by-step explanation:Solution:(0.2)x = 2.Taking log on both sideslog (0.2)x = log 2.x log (0.2) = 0.3010, [since log 2 = 0.3010].x log (2/10) = 0.3010.x [log 2 - log 10] = 0.3010.

Step-by-step explanation:Solution:(0.2)x = 2.Taking log on both sideslog (0.2)x = log 2.x log (0.2) = 0.3010, [since log 2 = 0.3010].x log (2/10) = 0.3010.x [log 2 - log 10] = 0.3010.x [log 2 - 1] = 0.3010,[since log 10=1].

Step-by-step explanation:Solution:(0.2)x = 2.Taking log on both sideslog (0.2)x = log 2.x log (0.2) = 0.3010, [since log 2 = 0.3010].x log (2/10) = 0.3010.x [log 2 - log 10] = 0.3010.x [log 2 - 1] = 0.3010,[since log 10=1].x [0.3010 -1] = 0.3010, [since log 2 = 0.3010].

Step-by-step explanation:Solution:(0.2)x = 2.Taking log on both sideslog (0.2)x = log 2.x log (0.2) = 0.3010, [since log 2 = 0.3010].x log (2/10) = 0.3010.x [log 2 - log 10] = 0.3010.x [log 2 - 1] = 0.3010,[since log 10=1].x [0.3010 -1] = 0.3010, [since log 2 = 0.3010].x[-0.699] = 0.3010.

Step-by-step explanation:Solution:(0.2)x = 2.Taking log on both sideslog (0.2)x = log 2.x log (0.2) = 0.3010, [since log 2 = 0.3010].x log (2/10) = 0.3010.x [log 2 - log 10] = 0.3010.x [log 2 - 1] = 0.3010,[since log 10=1].x [0.3010 -1] = 0.3010, [since log 2 = 0.3010].x[-0.699] = 0.3010.x = 0.3010/-0.699.

Step-by-step explanation:Solution:(0.2)x = 2.Taking log on both sideslog (0.2)x = log 2.x log (0.2) = 0.3010, [since log 2 = 0.3010].x log (2/10) = 0.3010.x [log 2 - log 10] = 0.3010.x [log 2 - 1] = 0.3010,[since log 10=1].x [0.3010 -1] = 0.3010, [since log 2 = 0.3010].x[-0.699] = 0.3010.x = 0.3010/-0.699.x = -0.4306….

Step-by-step explanation:Solution:(0.2)x = 2.Taking log on both sideslog (0.2)x = log 2.x log (0.2) = 0.3010, [since log 2 = 0.3010].x log (2/10) = 0.3010.x [log 2 - log 10] = 0.3010.x [log 2 - 1] = 0.3010,[since log 10=1].x [0.3010 -1] = 0.3010, [since log 2 = 0.3010].x[-0.699] = 0.3010.x = 0.3010/-0.699.x = -0.4306….x = -0.4 (nearest tenth)

Appropriate Question :-

If [tex]\rm \: {(0.2)}^{x} = 2 \\ \\ [/tex] and log2 = 0.3010, the the value of x to the nearest tenth is:

[tex] \\[/tex]

[tex]\large\underline{\sf{Solution-}}[/tex]

Given that,

[tex]\rm \: {(0.2)}^{x} = 2 \\ \\ [/tex]

can be rewritten as

[tex]\rm \: {\bigg(\dfrac{2}{10} \bigg) }^{x} = 2 \\ \\ [/tex]

On taking log on both sides, we get

[tex]\rm \: log {\bigg(\dfrac{2}{10} \bigg) }^{x} = log2 \\ \\ [/tex]

[tex]\rm \:x \: log {\bigg(\dfrac{2}{10} \bigg) } = log2 \\ \\ [/tex]

[tex]\rm \:x \: (log2 - log10) = log2 \\ \\ [/tex]

[tex]\rm \:x \: (0.3010 - 1) = 0.3010 \\ \\ [/tex]

[tex]\rm \: - 0.699x = 0.301 \\ \\ [/tex]

[tex]\rm \: x \: = \: - \: \dfrac{0.301}{0.699} \\ \\ [/tex]

[tex]\bf\implies \: x \: = \: \dfrac{301}{699} \\ \\ [/tex]

On taking log on both sides, we get

[tex]\rm \: logx = log\bigg(\dfrac{301}{699} \bigg) \\ \\ [/tex]

[tex]\rm \: logx = log301 - log699 \\ \\ [/tex]

[tex]\rm \: logx = 2.4786 - 2.8445 \\ \\ [/tex]

[tex]\rm \: logx = \: - \: 0.3659 \\ \\ [/tex]

[tex]\rm \: logx = \: - 1 + 1 - \: 0.3659 \\ \\ [/tex]

[tex]\rm \: logx = \: - 1 + 0.6341\\ \\ [/tex]

[tex]\rm \: logx = \: \overline{1} .6341\\ \\ [/tex]

[tex]\rm \: x = \: antilog( \overline{1} .6341)\\ \\ [/tex]

[tex]\rm \: x = {10}^{ - 1} \times 4.306 \\ \\ [/tex]

[tex]\rm \: x = 0.4306 \\ \\ [/tex]

[tex]\bf\implies \:x = 0.4 \\ \\ [/tex]

[tex]\rule{190pt}{2pt} \\ [/tex]

[tex] { \red{ \mathfrak{Additional\:Information}}}[/tex]

[tex]\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{ log_{x}(x) = 1}\\ \\ \bigstar \: \bf{ log_{x}( {x}^{y} ) = y}\\ \\ \bigstar \: \bf{ log_{ {x}^{z} }( {x}^{w} ) = \dfrac{w}{z} }\\ \\ \bigstar \: \bf{ log_{a}(b) = \dfrac{logb}{loga} }\\ \\ \bigstar \: \bf{ {e}^{logx} = x}\\ \\ \bigstar \: \bf{ {e}^{ylogx} = {x}^{y}}\\ \\ \bigstar \: \bf{log1 = 0}\\ \\ \end{array} }}\end{gathered}\end{gathered}\end{gathered}[/tex]