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solve for x
[tex] \huge(3 \sqrt{ \frac{3}{5} } )^{2x + 1} = \frac{125}{27} [/tex]
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[tex] \\ [/tex]
To solve the equation [tex] \huge(3 \sqrt{ \frac{3}{5} } )^{2x + 1} = \frac{125}{27} [/tex], we can start by simplifying the expression inside the parentheses.
The square root of [tex] \frac{3}{5} [/tex] can be written as [tex] \sqrt{ \frac{3}{5} } [/tex].
Now, let's rewrite the equation as [tex] \huge(3 \sqrt{ \frac{3}{5} } )^{2x + 1} = \frac{125}{27} [/tex].
Taking the square root of both sides, we have [tex] 3 \sqrt{ \frac{3}{5} } = \sqrt[2x+1]{ \frac{125}{27} } [/tex].
Now we can solve for x.
To solve the equation [tex] \huge(3 \sqrt{ \frac{3}{5} } )^{2x + 1} = \frac{125}{27} [/tex], we can start by simplifying the expression inside the parentheses.
The square root of [tex] \frac{3}{5} [/tex] can be written as [tex] \sqrt{ \frac{3}{5} } [/tex]. Simplifying this further, we get [tex] \frac{\sqrt{15}}{\sqrt{25}} [/tex], which simplifies to [tex] \frac{\sqrt{15}}{5} [/tex].
Now, let's rewrite the equation as [tex] \huge(3 \cdot \frac{\sqrt{15}}{5} )^{2x + 1} = \frac{125}{27} [/tex].
Simplifying the left side of the equation, we have [tex] \huge(\frac{3\sqrt{15}}{5})^{2x+1} [/tex].
To solve for x, we can take the logarithm of both sides of the equation. Taking the logarithm base 10, we get:
[tex] (2x+1) \log_{10}(\frac{3\sqrt{15}}{5}) = \log_{10}(\frac{125}{27}) [/tex].
Now we can solve for x by isolating it on one side of the equation.
Let's simplify the expression inside the parentheses. The square root of [tex] \frac{3}{5} [/tex] can be written as [tex] \sqrt{ \frac{3}{5} } [/tex]. Simplifying further, we get [tex] \frac{\sqrt{15}}{\sqrt{25}} [/tex], which is [tex] \frac{\sqrt{15}}{5} [/tex].
Now the equation becomes [tex] \huge(3 \cdot \frac{\sqrt{15}}{5} )^{2x + 1} = \frac{125}{27} [/tex].
To solve for x, we can take the logarithm of both sides of the equation. Taking the natural logarithm (ln) on both sides, we get:
[tex] (2x+1) \ln(3 \cdot \frac{\sqrt{15}}{5}) = \ln(\frac{125}{27}) [/tex].
Now we can isolate x by dividing both sides by [tex] 2 \ln(3 \cdot \frac{\sqrt{15}}{5}) [/tex]
The natural logarithm (ln) of [tex]3 \cdot \frac{\sqrt{15}}{5}[/tex] is approximately 0.530628. The natural logarithm (ln) of [tex]\frac{125}{27}[/tex] is approximately 1.321756.
So, the equation becomes:
[tex](2x + 1) \cdot 0.530628 = 1.321756[/tex].
To solve for x, we can divide both sides of the equation by 0.530628:
[tex]2x + 1 = \frac{1.321756}{0.530628}[/tex].
Now we can solve for x by subtracting 1 from both sides and dividing by 2:
[tex]x = \frac{\frac{1.321756}{0.530628} - 1}{2}[/tex].
Evaluating this expression, we find that x is approximately 0.482.
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