The Question is ...
By drawing a suitable straight line on the same axes , solve the equation
$11 = 7(2^x)$
How do I make this equation fit into the equation the graph of $y = 11/2^x + 5 $ ?
(I've already drawn the graph out)
I just have problems manipulating
$11= 7(2^x)$ to like example y= something so that I can draw the straight line onto my graph
Use:
- $$\ln(a^b)=b\ln(a)$$
- $$\ln\left(\frac{a}{b}\right)=\ln(a)-\ln(b)$$
$$11=7\cdot2^x\Longleftrightarrow$$ $$\frac{11}{7}=2^x\Longleftrightarrow$$ $$\ln\left(\frac{11}{7}\right)=x\ln(2)\Longleftrightarrow$$ $$x=\frac{\ln\left(\frac{11}{7}\right)}{\ln(2)}\Longleftrightarrow$$ $$x=\frac{\ln(11)-\ln(7)}{\ln(2)}$$
$$\frac{11}{7}=2^x \implies \log_{2}(11/7)=x \implies \frac{\log(11)-\log(7)}{\log(2)}=x$$
This question in particular is not going to straightforward without logarithms. Of course if you had some way of translating the base, you could proceed as follows:
$$(\frac{1}{5})^x=125^{x-3} \implies 5^{-x}=5^{3(x-3)} \implies -x=3x-9 \implies 4x=9$$
so $x=9/4$.
