Answered

At Westonci.ca, we connect you with the best answers from a community of experienced and knowledgeable individuals. Get detailed answers to your questions from a community of experts dedicated to providing accurate information. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.

2logbase4(x)-logbase4(5)=125

Sagot :

konrad509
[tex]D:x>0\\\\ 2\log_4x-\log_45=125\\ \log_4x^2-\log_45=\log_44^{125}\\ \log_4\dfrac{x^2}{5}=\log_44^{125}\\ \dfrac{x^2}{5}=4^{125}\\ x^2=5\cdot4^{125}=5\cdot2^{250}=5\cdot(2^{125})^2\\ x=\sqrt{5\cdot2^{250}} \vee x=-\sqrt{5\cdot2^{250}}\\ x=\sqrt{5\cdot4^{125}} \vee x=-\sqrt{5\cdot4^{125}}\\ x=\sqrt{5\cdot(2^{125})^2} \vee x=-\sqrt{5\cdot(2^{125})^2}\\ x=\sqrt5 \cdot\sqrt{(2^{125})^2} \vee x=-\sqrt5 \cdot\sqrt{(2^{125})^2} \\ x=2^{125}\sqrt5 \vee x=-2^{125}\sqrt5[/tex]

[tex]x=-2^{125}\sqrt5 \not \in D\Rightarrow \boxed{x=2^{125}\sqrt5}[/tex]
Thank you for your visit. We are dedicated to helping you find the information you need, whenever you need it. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Thank you for visiting Westonci.ca, your go-to source for reliable answers. Come back soon for more expert insights.