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

Quadratic equations and inequalities, variation equations, function notation, systems of equations, etc.
santaclaus
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### solving w/exponents: x^1.05 = 1.09, ln sqrt x + sqrt x = 3

How do I solve and equation like the following?

x1.05=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
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(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?

x1.05=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:

. . . . .$\log(x)\, =\, \frac{1.09}{1.05}$

. . . . .$10^{\log(x)}\, =\, 10^{\frac{1.09}{1.05}}$

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

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

stapel_eliz
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Joined: Mon Dec 08, 2008 4:22 pm
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If the instructions specify a need for a decimal approximation, then yes, you'd plug this into your calculator.