what is `2^(x+1) = 5`  using the logorithm?  must use the logorithm formula

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embizze's profile pic

embizze | High School Teacher | (Level 1) Educator Emeritus

Posted on

Solve `2^(x+1)=5` :

Take a logarithm of both sides. You can use any base -- typically you would choose to use base `e` (i.e. the natural logarithm) or base 10 (the common logarithm) as most scientific calculators can compute these.

Using the natural logarithm `ln` we get:

`ln(2^(x+1))=ln5`

A property of logarithms, of any base,is `log a^b=bloga` . So:

`(x+1)ln2=ln5`

`x+1=ln5/ln2`

`x=ln5/ln2 -1`

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The solution to `2^(x+1)=5` is `x=ln5/ln2-1~~1.322`

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Note that the answer provided by mjripalda is fundamentally the same -- you can use the change of base formula to convert `log_2 5` to `ln5/ln2` , or `log5/log2` , or any other base of your choosing.

Sources:
lemjay's profile pic

lemjay | High School Teacher | (Level 3) Senior Educator

Posted on

Note that an exponential expression has its equivalent logarithm form.

Exponential Expression     <====>    Logarithmic Expresion

         `b^m = a`                  <====>        `log_b a = m`  

 

So,

 `2^(x+1) = 5`    

It's equivalent logarithm form is,

`log_2 5 = x+1`          

Then, we may solve for x.

 `log_2 5 - 1= x`      

Answer:  x = `log_2 5`  -  1                       

                                                                  

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