Phoebe

2021-09-28

Solve the equation.
$\mathrm{log}\left(x+3\right)=1-\mathrm{log}\left(x-2\right)$

Talisha

Step 1
Given equation:
$\mathrm{log}\left(x+3\right)=1-\mathrm{log}\left(x-2\right)$
Solve the above equation, we get
$\mathrm{log}\left(x+3\right)+\mathrm{log}\left(x-2\right)=1$
Step 2
We know that $\mathrm{log}\left(x\right)+\mathrm{log}\left(y\right)=\mathrm{log}\left(x\cdot y\right)$
So, we have
$\mathrm{log}\left(x+3\right)+\mathrm{log}\left(x-2\right)=1$
$\mathrm{log}\left(x+3\right)\left(x-2\right)=1$
The value pf $\mathrm{log}\left(10\right)=1$
So,
$\mathrm{log}\left(x+3\right)\left(x-2\right)=\mathrm{log}\left(10\right)$
(x+3)(x-2)=10
${x}^{2}+x-6=10$
${x}^{2}+x-16=0$
Step 3
$x=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$
Here, a=1, b=1, and c=-16
Substitute these values, we get
$x=\frac{-1±\sqrt{{1}^{2}-4×1\left(-16\right)}}{2×1}$
$x=\frac{-1±\sqrt{1+64}}{2}$
$x=\frac{-1±\sqrt{65}}{2}$
Step 4
Take positive sign, we get
$x=\frac{-1+\sqrt{65}}{2}$
Take negative sign, we get
$x=\frac{-1+\sqrt{65}}{2}$
We take $x=\frac{-1+\sqrt{65}}{2}$ because the value of $x=\frac{-1+\sqrt{65}}{2}$ lead to logarithm negative.
Hence the value of $x=\frac{-1+\sqrt{65}}{2}$

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