Question

How do you solve `2 - log_3sqrt(x² +17) = 0`?

Answer 1

`x = +-8`

Explanation:

`2 = log_3(sqrt(x^2 + 17))`

`3^2 = sqrt(x^2 + 17)`

`9 = sqrt(x^2 + 17)`

`(9)^2 = (sqrt(x^2 + 17))^2`

`81 = x^2 + 17`

`81 - 17 = x^2`

`64 = x^2`

`+- 8 = x`

Slightly alternative approach:

`2 = log_3(sqrt(x^2 + 17))`

Use the rule `sqrt(x) = x^(1/2)`

`2 = log_3(x^2 + 17)^(1/2)`

Now use the rule `loga^n = nloga`

`2 = 1/2log_3(x^2 + 17)`

`2 /(1/2) = log_3(x^2 + 17)`

`4 = log_3(x^2 + 17)`

`3^4 = x^2 + 17`

`81 - 17 = x^2`

`64 = x^2`

`x = +-8`

Same answer!!

Hopefully this helps!

Answer 2

simplify using the properties of logarithms to find `x=+-8`

Explanation:

Lets start by putting the term with the `x` on one side of the equation and the constant on the other:

`log_3 sqrt(x^2+17) = 2`

The argument of the `log_3` can also be rewritten as a power

`log_3 [(x^2+17)^(1/2)] = 2`

We can then use the property of the logarithm that

`log a^b = b log a`

to rewrite our equation as

`1/2log_3 (x^2+17) = 2`

then multiplying both sides by `2` we get

`log_3 (x^2+17) = 4`

To get rid of the `log_3` we can put both sides of the equation to the power of `3`, remembering that `a^(log_a(b)) = b`, which gives us

`(x^2+17) = 3^4 = 81`

Subtracting `17` from both sides we get

`x^2 = 64`

and taking the square root we get

`x=+-8`