Algebra with Logarithm How do I solve:
$$-1.4= \log\left(\frac{x}{0.3}\right)$$
I've been out of school for 20 years and cannot remember order of operations of basic algebra!  please help!
 A: HINT:
Raise each side of the equation by a power of the base of your log (probably $2$ or $10$)...

You know what? Since you've been out of school for 20 years, I'll just cut you some slack and give you the full solution:


*

*$-1.4=\log_{2}\dfrac{x}{0.3} \implies 2^{-1.4}=\dfrac{x}{0.3} \implies 2^{-1.4}\cdot0.3=x \implies x\approx0.11367$

*$-1.4=\log_{10}\dfrac{x}{0.3} \implies 10^{-1.4}=\dfrac{x}{0.3} \implies 10^{-1.4}\cdot0.3=x \implies x\approx0.01194$
A: $$-1.4=\log_a{\frac{x}{0.3}}$$
Where $a$ is the base of the logarithm then by the definition of logarithm this is equivalent to:
$$a^{-1.4}=\frac{x}{0.3}$$
$$x=0.3a^{-1.4}$$
A: $$-1.4= \log(x/.3)$$
$$-14/10= \log(x/(3/10)$$
$$10^{-14/10}=x/(3/10)$$
$$10^{7/5}=10x/3$$
$$3\cdot10^{7/5}=10x$$
$$x=(3/10)\cdot10^{7/5}$$
$$x=3\cdot10^{7/5-1}$$
$$x=3\cdot10^{2/5}$$
A: First note that
$$ \log x=\log_{10} x $$
$$ \log\left(\frac{x}{y}\right)=\log x -\log y $$
$$ 10^{\log x}=x $$
So
$$ -1.4= \log\left(\frac{x}{0.3}\right)= \log x -\log 0.3 $$
$$ -1.4+\log 0.3= \log x $$
$$ 10^{-1.4+\log 0.3}=10^{\log x}=x $$
$$ x=10^{-1.4}\cdot 10^{\log 0.3}=0.3\cdot 10^{-1.4} $$
$$ x=0.0119432151166\dots $$
