Question

Use the change of base formula to approximate the solution to log0.5 15=1-2x

Answer 1

$\bf \textit{Logarithm Change of Base Rule} \\\\ log_a b\implies \cfrac{log_c b}{log_c a}\\\\ -------------------------------\\\\ log_{0.5}(15)=1-2x\implies \cfrac{log_{10}(15)}{log_{10}(0.5)}=1-2x \\\\\\ 2x=1-\cfrac{log_{10}(15)}{log_{10}(0.5)} \implies x=\cfrac{1-\frac{log_{10}(15)}{log_{10}(0.5)}}{2} \\\\\\ x=\cfrac{1}{2}-\cfrac{log_{10}(15)}{2log_{10}(0.5)}$

keeping in mind that for the change of base rule, it doesn't matter what the base "c" is, so long is the same above and below.

Answer 2

According to the Logarithm change of base formula,solution will be

             $\rm X=\dfrac{1}{2}-\dfrac{ \log_{10} 15}{\log_{10} 0.5}$

To solve the given question we need to understand change base rule of logarithm.

What is Logarithm Change of Base Rule?

         $\rm \log_{a} b= \dfrac{ \log_{c} b}{\log_{c} a} \;\;\;\;\;\;\;\;\;$  

where, we say Log b base a          

Given :   $\rm \log_{0.5} 15=1-2x$

               $\rightarrow\;\; \dfrac{\log_{10} 15}{\log_{10} 0.5}\\\\ \;\;\;\;\;\; \; \rm 1-2x$

               $\rm 2x=1-\dfrac{\log_{10} 15}{\log_{10} 0.5}\\\\x=\dfrac{1-\dfrac{\log_{10} 15}{\log_{10} 0.5}}{2}\\\\\\x=\dfrac{1}{2}-\dfrac{\log_{10} 15}{2 \log_{10} 0.5}$

Always remember Log base value will always be Positive.Log cannot be negative.

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