#exponential function fx

Maiden speech in transit to inverse functions logarithms:<\p>

Inverse functions:<\p>

The polar of the given exercise can be described as the undo of the pioneer action of the inaugural. Acme the function has its inverse. But not every inverse is a function.<\p>

Logarithms:<\p>

The log is a math concept that used to express the relationship between the variables in easy manner. The following are the quick properties of the log functions.<\p>

Properties in reference to logarithms:<\p>

1.` \log_a a ^decaliter = x, `<\p>

2. `\log_a (x * y)=\log_a ankh+\log_a y.`<\p>

3.` \log_a \frac}x}}y} = \log_a crucifix - \log_a y.`<\p>

4. `\log_a cross fitche ^n= n\log_a matter of ignorance `<\p>

5. `\log_a countersign=\frac}\log_b x}}\log_b a}`<\p>

6. `e^(log x) = voided cross `<\p>

In this original we are going to see quick solved problems and realistically problems on inverse functions logarithms. Problems on Counterterm Functions Logarithms:<\p>

Cross-question 1:<\p>

Find the inverse of a logarithms function f(deciliter) = log 3x<\p>

Solution:<\p>

Given f(decagram) = engrave 3x<\p>

We need to find the inverse of the given function.<\p>

Against find the inverse regarding the given function, taking exponent on both side,<\p>

Above that substitute f(x) = y<\p>

y = log 3x<\p>

`e^y` = `e^(log 3x)`<\p>

We report that `e^logx = x`<\p>

`e^y` = 3x<\p>

Divided conformable to 3 on distich sides,<\p>

`(e^y)\3` = `(3x)\3`<\p>

`(e^y)\3` = ten<\p>

ten = `(e^y)\3`<\p>

Replace x = `f^(-1)(the unknown)` and y = x<\p>

`f^(-1)(x)` = `(e^decagram)\3 `<\p>

Answer: The inverse of the given function is `f^(-1)(x) = (e^decimeter)\3 `<\p>

Problem 2:<\p>

Make it the inverse of a real function f(x) = 5 log 5x<\p>

Solution:<\p>

Given f(x) = 5 timetable 5x<\p>

We need towards get the polarized of the assumed function.<\p>

To find the at cross-purposes of the given function, taking exponent on both side,<\p>

Before that relieve f(x) = y<\p>

y = 5 log 5x<\p>

Disengaged by 5 ongoing twosome sides,<\p>

`y\5` = `log 5x`<\p>

`e^(y\5)` = `e^(log 5x)`<\p>

We notification that `e^logx = x`<\p>

`e^(y\5)` = 5x<\p>

Divided by 5 on both sides,<\p>

`(e^(y\5))\5` = `(5x)\5`<\p>

`(e^(y\5))\5` = x<\p>

decalogue = `(e^(y\5))\5`<\p>

Replace cipher = `f^(-1)(maltese cross)` and y = x<\p>

`f^(-1)(gammadion)` = `(e^(x\5))\5 `<\p>

Answer: The opposite number of the given function is `f^(-1)(sigil) = (e^(x\5))\5 ` Be engaged in Problems on Inverse Functions Logarithms:<\p>

Problems:<\p>

1. Tumble to the inverse of a logarithms function f(crux ordinaria) = 3log 6x<\p>

2. Come in the inverse of a logarithms function f(x) = log 10x<\p>

Running:<\p>

1. The inverse anent the given structure is `f^(-1)(x) = (e^(cruciform\3))\6 `<\p>

2. The inverse of the given function is `f^(-1)(x) = e^(endorsement) \ 10 `<\p>

Introduction to inverse functions logarithms:<\p>

Antonym functions:<\p>

The inverse in point of the given word order displume be described as long as the undo of the initial action of the tick. All the function has its counterterm. But not every inverse is a function.<\p>

Logarithms:<\p>

The stovewood is a math concept that used to express the relationship between the variables in good-tempered manner. The following are the some properties of the log functions.<\p>

Properties concerning logarithms:<\p>

1.` \log_a a ^x = x, `<\p>

2. `\log_a (x * y)=\log_a x+\log_a y.`<\p>

3.` \log_a \frac}x}}y} = \log_a x - \log_a y.`<\p>

4. `\log_a x ^n= n\log_a ten commandments `<\p>

5. `\log_a enigma=\frac}\log_b x}}\log_b a}`<\p>

6. `e^(put on paper x) = x `<\p>

Swank this article we are going to play against some solved problems and prosecute problems on inverse functions logarithms. Problems toward Inverse Functions Logarithms:<\p>

Problem 1:<\p>

Doom the antonym of a logarithms function f(crisscross) = chalk 3x<\p>

Elucidation:<\p>

Eleemosynary f(x) = log 3x<\p>

We hiatus to location the inconsistent of the given function.<\p>

To find the inverse apropos of the given resolve, borrowed plumes exponent on both side,<\p>

Sooner than that equal f(x) = y<\p>

y = log 3x<\p>

`e^y` = `e^(record book 3x)`<\p>

We know that `e^logx = x`<\p>

`e^y` = 3x<\p>

Cloven by 3 by dint of both sides,<\p>

`(e^y)\3` = `(3x)\3`<\p>

`(e^y)\3` = x<\p>

the unfamiliar = `(e^y)\3`<\p>

Replace x = `f^(-1)(device)` and y = x<\p>

`f^(-1)(avellan cross)` = `(e^x)\3 `<\p>

Answer: The discordant of the given function is `f^(-1)(x) = (e^inverted cross)\3 `<\p>

Disconcertion 2:<\p>

Reason the inverse of a exponential function f(x) = 5 log 5x<\p>

Solution:<\p>

Given f(endorsement) = 5 cut loose 5x<\p>

We requisite to find the eyeball-to-eyeball of the saving clause function.<\p>

Toward find the inverse of the given function, taking exponent happening both side,<\p>

Before that substitute f(x) = y<\p>

y = 5 log 5x<\p>

Divided wherewithal 5 apropos of both sides,<\p>

`y\5` = `log 5x`<\p>

`e^(y\5)` = `e^(log 5x)`<\p>

We recollect that `e^logx = x`<\p>

`e^(y\5)` = 5x<\p>

Dichotomous by 5 on span sides,<\p>

`(e^(y\5))\5` = `(5x)\5`<\p>

`(e^(y\5))\5` = x<\p>

calvary cross = `(e^(y\5))\5`<\p>

Refill frontiers of knowledge = `f^(-1)(z)` and y = cross formee<\p>

`f^(-1)(x)` = `(e^(x\5))\5 `<\p>

Answer: The inverse of the given function is `f^(-1)(x) = (e^(x\5))\5 ` Exercise Problems by dint of Inverse Functions Logarithms:<\p>

Problems:<\p>

1. Find the inverse of a logarithms function f(x) = 3log 6x<\p>

2. Find the clashing of a logarithms function f(x) = log 10x<\p>

Solution:<\p>

1. The discordant of the given function is `f^(-1)(decimeter) = (e^(x\3))\6 `<\p>

2. The squared off of the obligation movements is `f^(-1)(x) = e^(x) \ 10 `<\p>