spatularificw2

2022-09-09

Solve this equation ${4}^{{\mathrm{log}}_{2}(x)}-{2}^{{\mathrm{log}}_{2}(x)}={3}^{{\mathrm{log}}_{3}(12)}$

I thought to write ${2}^{{\mathrm{log}}_{2}(x{)}^{2}}-{2}^{{\mathrm{log}}_{2}(x)}={3}^{{\mathrm{log}}_{3}(12)}$. Then is there a way to factorize ${2}^{{\mathrm{log}}_{2}(x)}$? I don't know how to proceed...

I thought to write ${2}^{{\mathrm{log}}_{2}(x{)}^{2}}-{2}^{{\mathrm{log}}_{2}(x)}={3}^{{\mathrm{log}}_{3}(12)}$. Then is there a way to factorize ${2}^{{\mathrm{log}}_{2}(x)}$? I don't know how to proceed...

Helena Bentley

Beginner2022-09-10Added 10 answers

Let ${\mathrm{log}}_{2}x=t$, then ${2}^{{\mathrm{log}}_{2}(x)}={2}^{t}=x$. Thus the given equation is

$${x}^{2}-2x=12$$

$${x}^{2}-2x=12$$

Genesis Gibbs

Beginner2022-09-11Added 2 answers

Hint:

We have ${a}^{{\mathrm{log}}_{a}(b)}=b$ and we have the following:

$${4}^{{\mathrm{log}}_{2}(x)}={2}^{2{\mathrm{log}}_{2}(x)}=({2}^{{\mathrm{log}}_{2}(x)}{)}^{2}={x}^{2}$$

We have ${a}^{{\mathrm{log}}_{a}(b)}=b$ and we have the following:

$${4}^{{\mathrm{log}}_{2}(x)}={2}^{2{\mathrm{log}}_{2}(x)}=({2}^{{\mathrm{log}}_{2}(x)}{)}^{2}={x}^{2}$$

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