Question

What are the solutions to log6(x^2+8)= 1+ log 6 (x)

Answer 1

`\log_6(x^2+8)=1+\log_6x`

`\log_6(x^2+8)=\log_66+\log_6x`

`\log_6(x^2+8)=\log_6(6x)`

`6^(\log_6(x^2+8))=6^(\log_6(6x))`

`x^2+8=6x`

`x^2-6x+8=0`

`(x-4)(x-2)=0`

`\implies x=4,x=2`

Answer 2

x = 2 and x = 4

Explanation:

correct on edg, BUT IF YOU DON'T BELIEVE ME..... please, read below.

log6(x2+8)=1+log6(x)

Move all the terms containing a logarithm to the left side of the equation.

log6(x2+8)−log6(x)=1

Use the quotient property of logarithms, logb(x)−logb(y)=logb(xy)

log6(x2+8x)=1

Rewrite log6(x2+8x)=1

in exponential form using the definition of a logarithm. If x and b are positive real numbers and b≠1, then logb(x)=y is equivalent to by=x

61=x2+8x

Solve for x

Evaluate the exponent.

6=x2+8x

Rewrite the equation as x2+8x=61

x2+8x=6

Evaluate the exponent.

x2+8x=6

Solve for x

Multiply each term by x

and simplify.

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x2+8=6x

Subtract 6x

from both sides of the equation.

x2+8−6x=0

Factor x2+8−6x

using the AC method.

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(x−4)(x−2)=0

Set x−4

equal to 0 and solve for x

x=4

Set x−2

equal to 0 and solve for x

x=2

The solution is the result of x−4=0

and x−2=0

x=4,2