Question

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Answer 1

Answer:

An exponential function is a function of the form

f(x)=bx

where b≠1 is a positive real number. The domain of an exponential function is (−∞,∞) and the range is (0,∞).

Solve the equation: 52x−3=752x−3=7.

Since we can’t easily rewrite both sides as exponentials with the same base, we’ll use logarithms instead. Above we said that logb(x)=ylogb⁡(x)=y means that by=xby=x. That statement means that each exponential equation has an equivalent logarithmic form and vice-versa. We’ll convert to a logarithmic equation and solve from there.

52x−3log

⎛⎝⎜

⎞⎠⎟=7=2x−352x−3=7log5

⁡(7

)=2x−3

From here, we can solve for xx directly.

2xx=log5(7)+3=log5(7)+32

A logarithmic function is a function defined as follows

logb(x)=ymeans thatby=xlogb⁡(x)=ymeans thatby=x

where b≠1b≠1 is a positive real number. The domain of a logarithmic function is (0,∞)(0,∞) and the range is (−∞,∞)(−∞,∞).

Solve the equation:

log3(2x+1)=1−log3(x+2).log3⁡(2x+1)=1−log3⁡(x+2).

With more than one logarithm, we’ll typically try to use the properties of logarithms to combine them into a single term.

log3(2x+1)log3(2x+1)+log3(x+2)log3((2x+1)(x+2))log3(2x2+5x+2)2x2+5x+22x2+5x−1=1−log3(x+2)=1=1=1=3=0log3⁡(2x+1)=1−log3⁡(x+2)log3⁡(2x+1)+log3⁡(x+2)=1log3⁡((2x+1)(x+2))=1log3⁡(2x2+5x+2)=12x2+5x+2=32x2+5x−1=0

Let’s use quadratic formula to solve this.

x=−5±52−4⋅2⋅−1−−−−−−−−−−−√2⋅2=−5±

−−−−−−−−⎷4.x=−5±52−4⋅2⋅−12⋅2=−5±33

4.

What happens if we try to plug x=