5

Given that $\log 5 = 0.7$ (to one decimal place), find the value of $x$ such that $2^x = 10$ (again to one decimal place)

I don't know what to do with the information that $10^{0.7} = 5$. Why is this information useful?

Phaptitude
  • 2,249

3 Answers3

11

We know,$$\log_{10}10=1$$

But $$\log_{10}10=\log_{10}(2\cdot5)=\log_{10}2+\log_{10}5$$

$$\implies \log_{10}2=1-\log_{10}5=1-0.7=0.3$$

Now taking logarithm on the given equation $$x\log_{10}2=\log_{10}10=1$$ as $\log_a(b^m)=m\log_ab$

1

$2^x=10/\cdot \log_{10}$

$\log_{10}{2^x}=\log_{10}{10}$

$x\log_{10}{2}=1$

$x=\frac{1}{\log_{10}{2}}=\frac{1}{\log_{10}{\frac{10}{5}}}=\frac{1}{\log_{10}10-\log_{10}5}=\frac{1}{1-\log_{10}5}=\frac{1}{1-0.7}=\frac{1}{0.3}=\frac{1}{\frac{3}{10}}=\frac{10}{3}$

$x=\frac{10}{3}$

Madrit Zhaku
  • 5,294
0

$\log_a b = c$ represents $b=a^c$ by definition. Therefore, $2^x=10$ can be represented as $\log_2 10= x$. Thus $x=\log_2 10$. Period.

There are many properties in logarithm, one which is useful here is $\log_a b=\frac{\log_c b}{\log_c a}$. As a result, we can represent $x$ as

\begin{align*} x&= \log_2 10\\ &= \frac{\log_c 10}{\log_c 2} && c \text{ can be any positive number not equals to 1}\\ &= \frac{\log_{10} 10}{\log_{10} 2} && \text{based on the given question, we should choose } c=10\\ &= \frac{1}{0.3}\\ &=\frac{10}{3} \end{align*}