Averi Mitchell

2022-06-19

Solve $(lo{g}_{2}(x+1){)}^{2}=4$

$(lo{g}_{2}(x+1){)}^{2}=4$

$lo{g}_{2}(x+1)\ast lo{g}_{2}(x+1)=lo{g}_{2}16$

${x}^{2}+2x-15=0$

$(x+1)\ast (x+1)=16$

${x}^{2}+2x+1=16$

${x}^{2}+2x-15=0$

$(x+5)(x-3)=0$

${x}_{1}=-5;{x}_{2}=3$

The solution is only ${x}_{1}=3$. Is this correct?

$(lo{g}_{2}(x+1){)}^{2}=4$

$lo{g}_{2}(x+1)\ast lo{g}_{2}(x+1)=lo{g}_{2}16$

${x}^{2}+2x-15=0$

$(x+1)\ast (x+1)=16$

${x}^{2}+2x+1=16$

${x}^{2}+2x-15=0$

$(x+5)(x-3)=0$

${x}_{1}=-5;{x}_{2}=3$

The solution is only ${x}_{1}=3$. Is this correct?

Kaydence Washington

Beginner2022-06-20Added 32 answers

If

$({\mathrm{log}}_{2}(x+1){)}^{2}=4$

then

${\mathrm{log}}_{2}(x+1)={+2}$

or

${\mathrm{log}}_{2}(x+1)={-2}$

so $x+1={2}^{{+2}}$ or $x+1={2}^{{-2}}$

$({\mathrm{log}}_{2}(x+1){)}^{2}=4$

then

${\mathrm{log}}_{2}(x+1)={+2}$

or

${\mathrm{log}}_{2}(x+1)={-2}$

so $x+1={2}^{{+2}}$ or $x+1={2}^{{-2}}$

Lydia Carey

Beginner2022-06-21Added 9 answers

Your solution is incorrect. First note that

${\mathrm{log}}_{2}\left({a}^{2}\right)\ne {({\mathrm{log}}_{2}(a))}^{2}$

In your case, we have

${({\mathrm{log}}_{2}(x+1))}^{2}=4\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{\mathrm{log}}_{2}(x+1)=\pm 2\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}x+1={2}^{\pm 2}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}x+1=4\text{or}x+1={\displaystyle \frac{1}{4}}$

Hence,

$x=3\text{or}x=-{\displaystyle \frac{3}{4}}$

${\mathrm{log}}_{2}\left({a}^{2}\right)\ne {({\mathrm{log}}_{2}(a))}^{2}$

In your case, we have

${({\mathrm{log}}_{2}(x+1))}^{2}=4\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{\mathrm{log}}_{2}(x+1)=\pm 2\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}x+1={2}^{\pm 2}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}x+1=4\text{or}x+1={\displaystyle \frac{1}{4}}$

Hence,

$x=3\text{or}x=-{\displaystyle \frac{3}{4}}$