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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... ?

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    You can use the change-of-base formula and write $2 = \log_2x + \log_4x = \log_2 x + \log_2x/\log_24 = \log_2x + \frac12\log_2x = \frac32 \log_2x$. Now take it from here. – 2017-02-24

6 Answers 6

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$$\log_2 x +\log_4 x = 2$$ $$\frac{1}{\log_ x2} +\frac{1}{\log_ x4} = 2$$ $$\frac{1}{\log_ x2} +\frac{1}{2\log_ x2} = 2$$ $$\frac{3}{2\log_ x2} = 2$$ $$\frac{1}{\log_ x2} =\frac{4}{3}$$ $$\log_ 2x =\frac{4}{3}$$ $$x=2^{4/3}$$

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    Could you explain the last step? – 2017-02-24
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    @Melkor it is the definition of the logarithm. – 2017-02-24
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    Sorry I misread as a mixed fraction and got confused... – 2017-02-24
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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

$$2^{2\log_2(x)}4^{\log_4(x)}=16$$

$$x^2\cdot x=16$$

$$x^3=16$$

$$x=\sqrt[3]{16}\approx2.7$$

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    Could you explain how you got to $x^2 * x =16$ ? – 2017-02-24
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    Note the logarithms cancel by definition, since: $$4^{\log_4(x)}=x$$and$$2^{2\log_2(x)}=(2^{\log_2(x)})^2=x^2$$ – 2017-02-24
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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?

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You would need to change the base of one of the $log$. For example $\log_4 x = \frac{\log_2 x}{log_2 4} = \frac{\log_2 x}{2}$ $$ 2 = \log_2 x + \log_4 x = \frac{3}{2} \log_2x \implies x = 2^{4/3} = \sqrt[3]{16} $$

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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 $x=\exp(2/3)$.

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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.