Lisantiom

2022-10-07

A Simple Logarithm Question

Solve for $x$: ${\mathrm{log}}_{2}(2x+8)=3$

Correct Solution:

$2x+8={2}^{3}$

$2x+8=8$

$2x=0$

$x=0$

Why doesn't this work:

${\mathrm{log}}_{2}(2x+8)=3$

Expand:

${\mathrm{log}}_{2}(2x)+{\mathrm{log}}_{2}8=3$

${\mathrm{log}}_{2}(2x)+3=3$

${\mathrm{log}}_{2}(2x)=0$

$2x={2}^{0}$

$2x=1$

$x=1/2$

Thank you

Solve for $x$: ${\mathrm{log}}_{2}(2x+8)=3$

Correct Solution:

$2x+8={2}^{3}$

$2x+8=8$

$2x=0$

$x=0$

Why doesn't this work:

${\mathrm{log}}_{2}(2x+8)=3$

Expand:

${\mathrm{log}}_{2}(2x)+{\mathrm{log}}_{2}8=3$

${\mathrm{log}}_{2}(2x)+3=3$

${\mathrm{log}}_{2}(2x)=0$

$2x={2}^{0}$

$2x=1$

$x=1/2$

Thank you

Frederick Espinoza

Beginner2022-10-08Added 7 answers

Note that

$${\mathrm{log}}_{2}(2x+8)\ne {\mathrm{log}}_{2}(2x)+{\mathrm{log}}_{2}(8).$$

We have ${\mathrm{log}}_{2}(2x)+{\mathrm{log}}_{2}(8)={\mathrm{log}}_{2}(2x{\times}8)$

$${\mathrm{log}}_{2}(2x+8)\ne {\mathrm{log}}_{2}(2x)+{\mathrm{log}}_{2}(8).$$

We have ${\mathrm{log}}_{2}(2x)+{\mathrm{log}}_{2}(8)={\mathrm{log}}_{2}(2x{\times}8)$

Ratuiszt

Beginner2022-10-09Added 1 answers

One more way: to make yr life easier use the property of yhe logsrithm:

$${\mathrm{log}}_{a}b=\frac{\mathrm{ln}b}{\mathrm{ln}a}$$

wher3 ln is a natural log. After a bit of algebra exponentiate both sides.

$${\mathrm{log}}_{a}b=\frac{\mathrm{ln}b}{\mathrm{ln}a}$$

wher3 ln is a natural log. After a bit of algebra exponentiate both sides.

$\frac{20b}{{\left(4{b}^{3}\right)}^{3}}$

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