Question

If (0.2)x = 2 and log 2 = 0.3010, then the value of x to the nearest tenth is: (a) -10.0, (b) -0.5, (c) -0.4, (d) -0.2, (e) 10.0​

Answer 1

Appropriate Question :-

$\sf \: If \: {(0.2)}^{x} = 2 \: and \: log 2 = 0.3010,$

then the value of x to the nearest tenth is

(a) -10.0

(b) -0.5

(c) -0.4

(d) -0.2

(e) 10.0

$\large\underline{\sf{Solution-}}$

Given that,

$\rm \: log2 = 0.3010$

And

$\rm \: {(0.2)}^{x} = 2$

can be rewritten as

$\rm \: {\bigg(\dfrac{2}{10} \bigg) }^{x} = 2$

On taking log on both sides, we get

$\rm \: log {\bigg(\dfrac{2}{10} \bigg) }^{x} = log2$

We know,

$\boxed{\sf{ \:log {x}^{y} = ylogx \: }}$

So, using this identity, we get

$\rm \: x \: log\bigg(\dfrac{2}{10} \bigg) = log2$

We know,

$\boxed{\sf{ \:log \bigg(\frac{x}{y}\bigg) = logx - logy \: }}$

So, using this, we get

$\rm \: x(log2 - log10) = log2$

As it is given that, log 2 = 0.3010 and we know that log10 = 1

So, on substituting these values, we get

$\rm \: x(0.3010 - 1) = 0.3010$

$\rm \: - 0.699x = 0.3010$

$\rm \: x = - \dfrac{0.3010}{0.699}$

$\bf\implies \:x \: = \: - \: 0.4 \: approx$

[ as we have to find the value of x to nearest tenth ]

Hence,

Option (c) is correct.

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Additional Information :-

$\boxed{\sf{ \:logxy = logx + logy \: }}$

$\boxed{\sf{ \: log_{a}(b) = \frac{logb}{loga} \: }}$

$\boxed{\sf{ \: log_{a}(a) = 1 \: }}$

$\boxed{\sf{ \: log_{a}( {a}^{x} ) = x \: }}$

$\boxed{\sf{ \: log_{ {a}^{y} }( {a}^{x} ) = \frac{x}{y} \: }}$

$\boxed{\sf{ \: log_{ {b}^{y} }( {a}^{x} ) = \frac{x}{y} log_{b}(a)\: }}$

$\boxed{\sf{ \: {a}^{ log_{a}(x)} = x \: }}$

$\boxed{\sf{ \: {a}^{ y \: log_{a}(x)} = {x}^{y} \: }}$

Answer 2

$\large\mathbb \red{\fcolorbox{red}{lime}{ answer \ }}$