Inverse logarithms In an A level C2 textbook I have found a question:
$\log_2 x +\log_4 x = 2$
From attempting the question, I suspect I may need to try and inverse the logarithm, and on testing, I found that e to the power of ... looked promising but I am not sure. And my knowledge of logarithms is very limited.
Can anyone solve this and explain the answer, and also, because I think it is closely linked to the question, why is e useful and why is it 2.7... ?
 A: $$\log_2 x +\log_4 x = 2$$
$$\frac{1}{\log_ x2} +\frac{1}{\log_ x4} = 2$$
$$\frac{1}{\log_ x2} +\frac{1}{2\log_ x2} = 2$$
$$\frac{3}{2\log_ x2} = 2$$
$$\frac{1}{\log_ x2} =\frac{4}{3}$$
$$\log_ 2x =\frac{4}{3}$$
$$x=2^{4/3}$$
A: If we apply 4^ to both sides, we get
$$4^{\log_2(x)+\log_4(x)}=16$$
$$4^{\log_2(x)}4^{\log_4(x)}=16$$
Since $4=2^2$, this reduces to
$$2^{2\log_2(x)}4^{\log_4(x)}=16$$
$$x^2\cdot x=16$$
$$x^3=16$$
$$x=\sqrt[3]{16}\approx2.7$$
A: we write $$\frac{\ln(x)}{\ln(2)}+\frac{\ln(x)}{2\ln(2)}=2$$ multiplying by $$2\ln(2)$$ we obtain
$$2\ln(x)+\ln(x)=4\ln(2)$$thus we get
$$\ln(x)=\frac{4}{3}\ln(2)$$ can you finish this?
A: You would need to change the base of one of the $log$. 
For example $\log_4 x = \frac{\log_2 x}{log_2 4} = \frac{\log_2 x}{2}$
$$
2 = \log_2 x + \log_4 x = \frac{3}{2} \log_2x \implies x = 2^{4/3} = \sqrt[3]{16}
$$
A: Use the definition of $\log_a x=\ln(x)/\ln(a)$ and multiply $\log_2 x + \log_4 x = 2$ by $2 \ln 2$ to get $3\ln x=2$, hence $x=\exp(2/3)$.
A: Here is another way to think.  First note that $4$ is a power of $2$
Let $\log_4x=k\iff 4^k=x$
Now $4^k=x \implies (2^2)^k=x \implies 2^{2k}=x\iff\log_2x=2k\implies \frac{1}{2}\log_2{x}=\log_4{x}$
We now have $$\log_2x+\frac{1}{2}\log_2x=2\\\frac{3}{2}\log_2x=2\\\log_2x^{3/2}=2\iff2^2=x^{3/2}\implies x=16^{1/3}$$
I added this because it reinforces the bonds between logs and powers.
