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=
= 0.22)