Question

20" id="TexFormula1" title="4 {}^{x } + 6 {}^{x} = 9 {}^{x} " alt="4 {}^{x } + 6 {}^{x} = 9 {}^{x} " align="absmiddle" class="latex-formula">
please find x. and with explanation thanks ​

Answer 1

So, the value of x is 1.19

The question is an exponential equation

What is an exponential equation?

An exponential equation is a mathematical expression between two quantities in which one variable is raised to a power of the other variable.

How to find x?

Since $4^{x} + 6^{x} = 9^{x}$,

Dividing through by 4ˣ, we have

$\frac{4^{x} }{4^{x} } + \frac{6^{x} }{4^{x} } = \frac{9^{x} }{4^{x} } \\1 + (\frac{6}{4})^{x} } = (\frac{9}{4})^{x} } \\1 + (\frac{3}{2})^{x} } = (\frac{3^{2} }{2^{2} })^{x} } \\1 + (\frac{3}{2})^{x} } = (\frac{3}{2})^{2x} }$

Let y = (3/2)ˣ

So,

1 + y = y²

y² - y - 1 = 0

Using the quadratic formula to find y,

$y = \frac{-b +/- \sqrt{b^{2} - 4ac} }{2a}$

where a = 1 b = -1 and c = -1

Substituting the values of the variables into the equation, we have

$y = \frac{-(-1) +/- \sqrt{(-1)^{2} - 4\times 1 \times (-1)} }{2\times 1}\\= \frac{1 +/- \sqrt{1 + 4} }{2}\\= \frac{1 +/- \sqrt{5} }{2}\\= \frac{1 - \sqrt{5} }{2} or \frac{1 + \sqrt{5} }{2}\\= \frac{1 - 2.236}{2} or \frac{1 + 2.236}{2}\\= \frac{- 1.236}{2} or \frac{3.236}{2}\\= -0.618 or 1.618$

Since y = (3/2)ˣ

Takung logarithm of both sides, we have

㏒y = ㏒(3/2)ˣ

㏒y = x㏒(3/2)

x = ㏒y/㏒(3/2)

x = ㏒y/㏒1.5

Since we do not have logarithm of a negative number, we use y = 1.618.

So, x = ㏒y/㏒1.5

x = ㏒1.618/㏒1.5

x = 0.2090/0.1761

x = 1.19

So, the value of x is 1.19

Learn more about exponential equation here:

brainly.com/question/11832081

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