Phoebe

2021-09-28

Solve the equation.

$\mathrm{log}(x+3)=1-\mathrm{log}(x-2)$

Talisha

Skilled2021-09-29Added 93 answers

Step 1

Given equation:

$\mathrm{log}(x+3)=1-\mathrm{log}(x-2)$

Solve the above equation, we get

$\mathrm{log}(x+3)+\mathrm{log}(x-2)=1$

Step 2

We know that$\mathrm{log}\left(x\right)+\mathrm{log}\left(y\right)=\mathrm{log}(x\cdot y)$

So, we have

$\mathrm{log}(x+3)+\mathrm{log}(x-2)=1$

$\mathrm{log}(x+3)(x-2)=1$

The value pf$\mathrm{log}\left(10\right)=1$

So,

$\mathrm{log}(x+3)(x-2)=\mathrm{log}\left(10\right)$

(x+3)(x-2)=10

${x}^{2}+x-6=10$

${x}^{2}+x-16=0$

Step 3

Apply the quadratic formula

$x=\frac{-b\pm \sqrt{{b}^{2}-4ac}}{2a}$

Here, a=1, b=1, and c=-16

Substitute these values, we get

$x=\frac{-1\pm \sqrt{{1}^{2}-4\times 1(-16)}}{2\times 1}$

$x=\frac{-1\pm \sqrt{1+64}}{2}$

$x=\frac{-1\pm \sqrt{65}}{2}$

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$x=\frac{-1+\sqrt{65}}{2}$

Given equation:

Solve the above equation, we get

Step 2

We know that

So, we have

The value pf

So,

(x+3)(x-2)=10

Step 3

Apply the quadratic formula

Here, a=1, b=1, and c=-16

Substitute these values, we get

Step 4

Take positive sign, we get

Take negative sign, we get

We take

Hence the value of

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

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