solving w/exponents: x^1.05 = 1.09, ln sqrt x + sqrt x = 3

[santaclaus]

How do I solve and equation like the following?

x^1.05=1.09?
I would first do 1.05 log x = log 9
then, log x = log 1.09/1.05 is this correct?

the next one:

In sq.rt x+sq.rt.x = 3 (again, don't know notation for square root) here there are 2 square roots in the problem.

[stapel_eliz]

(again, don't know notation for square root)

Please review the article on formatting math as text, or else follow the instructions in the forum posting on LaTeX formatting.

How do I solve and equation like the following?

x^1.05=1.09?
I would first do 1.05 log x = log 9

I will guess that you are using the common (base-ten) log, but I'm afraid I don't follow your steps...?
To learn the properties of logs, try here. But I think your only error is that you changed the "9" to a "1.09", or vice versa.

Then next step in solving the log equation is to isolate the x-term, and then convert back to exponentials:

. . . . .

. . . . .

Simplify the left-hand side to just "x", and use your calculator if you need a decimal approximation for the right-hand side.

In sq.rt x+sq.rt.x = 3 here there are 2 square roots in the problem.

There may be two square roots, but unfortunately you haven't used grouping symbols to say where they go. Do you mean either of the following?

. . . . .ln(sqrt(x)) + sqrt(x) = 3

. . . . .ln(sqrt(x) + sqrt(x)) = 3

When you reply, please provide the instructions for this equation, along with a clear listing of everything you have tried so far. Thank you!

[santaclaus]

for the answer log 1.05/log 1.09, do I just put them into the calculator and divide?

[stapel_eliz]

If the instructions specify a need for a decimal approximation, then yes, you'd plug this into your calculator.