Need help explaining why this happens

Mathematics

1 answer:

Charra [1.4K]2 years ago

8 0

Just like 2 squared = 4, the square root of 2 multiplied by itself will result in a product of 2. Squaring a number always means multiplying that number by the same number.

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Rewrite the expression 7x+21=

Dima020 [189]

Minus the 21 from the other side of the equation making it 7x=-21 then u divide the 7 from the x to the -21 resulting in x=-3

5 0

2 years ago

Read 2 more answers

5^(-x)+7=2x+4 This was on plato

Setler79 [48]

Answer:

Below

I hope its not too complicated

$x=\frac{\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)}{\ln \left(5\right)}+\frac{3}{2}$

Step-by-step explanation:

$5^{\left(-x\right)}+7=2x+4\\\\\mathrm{Prepare}\:5^{\left(-x\right)}+7=2x+4\:\mathrm{for\:Lambert\:form}:\quad 1=\left(2x-3\right)e^{\ln \left(5\right)x}\\\\\mathrm{Rewrite\:the\:equation\:with\:}\\\left(x-\frac{3}{2}\right)\ln \left(5\right)=u\mathrm{\:and\:}x=\frac{u}{\ln \left(5\right)}+\frac{3}{2}\\\\1=\left(2\left(\frac{u}{\ln \left(5\right)}+\frac{3}{2}\right)-3\right)e^{\ln \left(5\right)\left(\frac{u}{\ln \left(5\right)}+\frac{3}{2}\right)}$

$Simplify\\\\\mathrm{Rewrite}\:1=\frac{2e^{u+\frac{3}{2}\ln \left(5\right)}u}{\ln \left(5\right)}\:\\\\\mathrm{in\:Lambert\:form}:\quad \frac{e^{\frac{2u+3\ln \left(5\right)}{2}}u}{e^{\frac{3\ln \left(5\right)}{2}}}=\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}$

$\mathrm{Solve\:}\:\frac{e^{\frac{2u+3\ln \left(5\right)}{2}}u}{e^{\frac{3\ln \left(5\right)}{2}}}=\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}:\quad u=\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)\\\\\mathrm{Substitute\:back}\:u=\left(x-\frac{3}{2}\right)\ln \left(5\right),\:\mathrm{solve\:for}\:x$

$\mathrm{Solve\:}\:\left(x-\frac{3}{2}\right)\ln \left(5\right)=\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right):\\\quad x=\frac{\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)}{\ln \left(5\right)}+\frac{3}{2}$