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

2 Answers
Jun 2, 2016

Answer:

#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!

Jun 2, 2016

Answer:

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#