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Let x be a real number such that 625^x = 64.

 

Then  125^x = \(?\sqrt?\)

 Apr 5, 2023
 #1
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We can use the fact that $625=5^4$ and $64=2^6$ to rewrite the given equation as:

(5^4)^x = 2^6

Simplifying the left-hand side using the power of a power rule gives:

5^(4x) = 2^6

We can solve for $x$ by taking the logarithm of both sides with respect to 4, 2 or any other base:

\begin{align*} 4x\log_5(5) &= 6\log_5(2) \ 4x &= \frac{6}{\log_5(5) - \log_5(2)} \ 4x &= \frac{6}{1 - \log_2(5)} \ x &= \frac{3}{2(1-\log_2(5))} \end{align*}

Now, to find $125^x$, we can use the fact that $125=5^3$ and substitute the value we found for $x$:

125^x = (5^3)^(3/(2(1 - log_2(5))) = 5^(9/2(1 - log_2(5)))

Therefore, the value of 125^x is:

5^(9/2(1 - log_2(5))) = 10*sqrt(5)

 Apr 5, 2023
 #2
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\(If \quad 625^x=64\qquad then\;find\qquad 125^x\\~\\ 625=5^4\\ 125=5^3\\ 64=4^3=2^6\\~\\ 625^x=64\\ 5^{3x\cdot \frac{4}{3}}=64\\ (125^x)^{\frac{4}{3}}=4^3\\ (125^x)=4^{3 \cdot \frac{3}{4}}\\ 125^x=4^{\frac{9}{4}}\\ 125^x=4^2*4^{1/4}\\ 125^x=4^2*2^{1/2}\\ 125^x=16\sqrt2 \)

 

NOTE:  I could easily have made a careless error.

 

 

LaTex

If \quad 625^x=64\qquad then\;find\qquad 125^x\\~\\
625=5^4\\
125=5^3\\
64=4^3=2^6\\~\\
625^x=64\\
5^{3x\cdot \frac{4}{3}}=64\\
(125^x)^{\frac{4}{3}}=4^3\\
(125^x)=4^{3 \cdot \frac{3}{4}}\\
125^x=4^{\frac{9}{4}}\\
125^x=4^2*4^{1/4}\\
125^x=4^2*2^{1/2}\\
125^x=16\sqrt2

 Oct 24, 2023

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