Condense them to the same base before solving for x
$\mathrm{log}16\left(x\right)+\mathrm{log}4\left(x\right)+\mathrm{log}2\left(x\right)=7$

Aneeka Hunt
2020-10-20
Answered

Condense them to the same base before solving for x
$\mathrm{log}16\left(x\right)+\mathrm{log}4\left(x\right)+\mathrm{log}2\left(x\right)=7$

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Brighton

Answered 2020-10-21
Author has **103** answers

so you hav

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Logarithmus to simple subtraction - how?

I am learning for a math exam and have the following solution:

$0.01={0.5}^{n}\phantom{\rule{0ex}{0ex}}n\cdot \mathrm{log}0.5=\mathrm{log}0.01\phantom{\rule{0ex}{0ex}}n=\frac{\mathrm{log}0.01}{\mathrm{log}0.5}$

OK, so far, so good. (I guess)

But now, it gets weird:

$n=\frac{\mathrm{log}0.01}{\mathrm{log}0.5}=\frac{0-2}{0.7-1}=\dots $

Can somebody please explain how to go from $\mathrm{log}0.01$ to $0-2$ and from $\mathrm{log}0.5$ to $0.7-1$

I am learning for a math exam and have the following solution:

$0.01={0.5}^{n}\phantom{\rule{0ex}{0ex}}n\cdot \mathrm{log}0.5=\mathrm{log}0.01\phantom{\rule{0ex}{0ex}}n=\frac{\mathrm{log}0.01}{\mathrm{log}0.5}$

OK, so far, so good. (I guess)

But now, it gets weird:

$n=\frac{\mathrm{log}0.01}{\mathrm{log}0.5}=\frac{0-2}{0.7-1}=\dots $

Can somebody please explain how to go from $\mathrm{log}0.01$ to $0-2$ and from $\mathrm{log}0.5$ to $0.7-1$

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What are the Limits at Infinity of Rational Functions?