Stuck on this logarithms question

Question

How do I solve this?

5^𝑥/√5 = 2(6^𝑥2−2)

I need to round to three decimals wherever it's needed too.

Answer 1

Rewrite the left side as 5^(x-1/2).

5^(x-1/2)=2(6^{(x^2-2)})

Since it does not look possible to make the powers have the same integer base, take the natural log of both sides:

ln[5^(x-1/2)] = ln[2(6^{(x^2-2)}]

Using properties of logarithms:

(x-1/2)ln5 = ln2 + (x²-2)ln6

In order to collect the terms containing x, use distributive property to remove parenthesis:

xln5 -(1/2)ln5 = ln2 + x²ln6-2ln6

This is quadratic in x so set equal to zero and use the quadratic formula:

x²ln6-xln5+ln2+(1/2)ln5-2ln6 = 0

So:

a = ln6

b=-ln5

c = ln2 + (1/2)ln5-2ln6

You can use properties of logarithms to simplify the expression for c, but this is not necessary:

c = ln[2(√5)(1/36) = ln(√5/18)

Take it from here using the quadratic formula.

Check your work here:

desmos.com/calculator/gxkakrdefd