The following geometric series has a sum to infinity as shown: 16log_(sqrtx)64+8log_(sqrtx)64+4log_(sqrtx)64+dots=(64)/(3). Calculate the value of x.

Question

The following geometric series has a sum to infinity as shown:  16log_(sqrtx)64+8log_(sqrtx)64+4log_(sqrtx)64+dots=(64)/(3). Calculate the value of  x.

Bytelearn · Accepted Answer

Identify Terms and Ratio: Identify the first term (a) and the common ratio (r) of the geometric series. The first term is $16\log_{\sqrt{x}}64$, and each subsequent term is half the previous term, which means the common ratio $r$ is $\frac{1}{2}$.Use Sum Formula: Use the formula for the sum to infinity of a geometric series, which is $S_{\infty} = \frac{a}{1 - r}$, where $S_{\infty}$ is the sum to infinity. Given that $S_{\infty} = \frac{64}{3}$, we can set up the equation $\frac{64}{3} = \frac{16\log_{\sqrt{x}}64}{1 - \frac{1}{2}}$.Solve for First Term: Solve for the first term $a$ by simplifying the denominator of the right side of the equation. Since $1 - \frac{1}{2} = \frac{1}{2}$, the equation becomes $\frac{64}{3} = \frac{16\log_{\sqrt{x}}64}{\frac{1}{2}}$.Multiply and Calculate: Multiply both sides of the equation by $\frac{1}{2}$ to isolate the first term $a$. This gives us $16\log_{\sqrt{x}}64 = \frac{64}{3} \times \frac{1}{2}$.Divide and Solve Log: Calculate the right side of the equation. $\frac{64}{3} \times \frac{1}{2} = \frac{64}{6} = \frac{32}{3}$. So, $16\log_{\sqrt{x}}64 = \frac{32}{3}$.Convert to Exponential: Divide both sides of the equation by 16 to solve for $\log_{\sqrt{x}}64$. $\log_{\sqrt{x}}64 = \frac{32}{3} \times \frac{1}{16}$.Recognize and Rewrite: Calculate the right side of the equation. $\frac{32}{3} \times \frac{1}{16} = \frac{32}{48} = \frac{2}{3}$. So, $\log_{\sqrt{x}}64 = \frac{2}{3}$.Simplify Exponent: Convert the logarithmic equation to its exponential form. $\sqrt{x}^{\frac{2}{3}} = 64$.Raise to Power: Recognize that $64$ is $2^6$, so the equation becomes $\sqrt{x}^{\frac{2}{3}} = 2^6$.Calculate Value: Since $\sqrt{x}$ is $x^{\frac{1}{2}}$, rewrite the equation as $x^{\frac{1}{2} \cdot \frac{2}{3}} = 2^6$.Simplify the exponent on the left side of the equation. $\frac{1}{2} \cdot \frac{2}{3} = \frac{1}{3}$, so the equation becomes $x^{\frac{1}{3}} = 2^6$.Raise both sides of the equation to the power of 3 to solve for $x$. $x = (2^6)^3$.Calculate the value of $x$. $x = 2^{18}$.Recognize that $2^{18}$ is a large number, but it is the correct value for $x$ in this context.

The following geometric series has a sum to infinity as shown: $\newline$ $16 \log _{\sqrt{x}} 64+8 \log _{\sqrt{x}} 64+4 \log _{\sqrt{x}} 64+\ldots=\frac{64}{3} .$ $\newline$ Calculate the value of $x$ .

Full solution

More problems from Sum of finite series not start from 1

Related topics

The following geometric series has a sum to infinity as shown:\newline16log⁡x64+8log⁡x64+4log⁡x64+…=643. 16 \log _{\sqrt{x}} 64+8 \log _{\sqrt{x}} 64+4 \log _{\sqrt{x}} 64+\ldots=\frac{64}{3} . 16logx​​64+8logx​​64+4logx​​64+…=364​.\newlineCalculate the value of x x x.

The following geometric series has a sum to infinity as shown: $\newline$ $16 \log _{\sqrt{x}} 64+8 \log _{\sqrt{x}} 64+4 \log _{\sqrt{x}} 64+\ldots=\frac{64}{3} .$ $\newline$ Calculate the value of $x$ .