# 3 ways to solve equations with unknown exponent (s)

Equations containing powers are sometimes a little tricky to solve. In any case, they require to know perfectly the rules of resolution. You will perhaps one day be asked to solve equations with somewhat original powers, in that the unknown will be exponent. If you have powers relating to the same base, the problem is easy to solve. If the bases are different, you will need to use the logarithm function… and a scientific calculator.

## Steps

### Method 1 of 3: Find an unknown by superscript (same basis)

#### Step 1. Check that the two powers have the same base

The base of a power is the value raised to the power, it is of the two values ​​that which is written in large. This method can only be used if there is an equality of powers both having the same base, the unknown being an exponent.

• The equation 65 + y = 63 { displaystyle 6 ^ {5 + y} = 6 ^ {3}}

comporte bien, comme le voyez, deux puissances ayant la même base, à savoir 6.

#### Step 2. To solve the equation, ignore the base

The powers being equal and having the same base, it is because their exponents are equal. That is why you can kind of ignore the base and enter the equality of the exponents.

• In the equation 65 + y = 63 { displaystyle 6 ^ {5 + y} = 6 ^ {3}}

, les puissances étant égales et ayant la même base, vous pouvez poser l'équation suivante: 5+y=3{displaystyle 5+y=3}

#### Step 3. Solve the equation

First, isolate the unknown on one side. As a reminder, any operation performed on one member of an equation must be done identically on the other member.

• Here, to isolate y { displaystyle y}

à gauche, vous allez soustraire 5 de chaque côté:

5+y=3{displaystyle 5+y=3}

5+y−5=3−5{displaystyle 5+y-5=3-5}

y=−2{displaystyle y=-2}

(solution).

#### Step 4. Check your work

The verification consists in taking the result found and making the numerical application in the starting equation, that is to say in replacing the unknown by the solution found. After calculations, equality must be verified. If it wasn't, double check, and if you still come across an inequality, go back to your calculations.

• You found that: y = −2 { displaystyle y = -2}

. Dans l'équation de départ, remplacez y{displaystyle y}

par -2 et faites les calculs:

65+y=63{displaystyle 6^{5+y}=6^{3}}

65−2=63{displaystyle 6^{5-2}=6^{3}}

63=63{displaystyle 6^{3}=6^{3}}

### Méthode 2 sur 3: Résoudre une équation avec l'inconnue en exposant

#### Step 1. Isolate the power

We will take the case of an equation with a power equal to a numerical constant (an integer, for example). If not already, isolate the power on the left and the constants on the right.

• Suppose you have to solve 3x − 5−2 = 79 { displaystyle 3 ^ {x-5} -2 = 79}

. Gardez la puissance (3x−5{displaystyle 3^{x-5}}

) à gauche et placez toutes les constantes à droite. Pour cela, ajoutez 2 de chaque côté de l'équation, ce qui donne:

3x−5−2=79{displaystyle 3^{x-5}-2=79}

3x−5−2+2=79+2{displaystyle 3^{x-5}-2+2=79+2}

3x−5=81{displaystyle 3^{x-5}=81}

#### Step 2. Rewrite the equation

The objective is to achieve equality between two powers having the same base. It is therefore advisable to transform the constant into a power. If you are unable to do so, then this method is not applicable to you.

• Let's go back to our equation 3x − 5 = 81 { displaystyle 3 ^ {x-5} = 81}

. Il faut donc transformer 81 en une puissance de base 3 pour résoudre simplement l'équation. Par chance (mais c'était prévu !), 81 est la puissance quatrième de 3, puisque 81=3×3×3×3=34{displaystyle 81=3\times 3\times 3\times 3=3^{4}}

. L'équation devient alors: 3x−5=34{displaystyle 3^{x-5}=3^{4}}

#### Step 3. Put the exponents even

The constant could be transformed into a power, you now have two equal powers with the same base. Ignore the base in question and equalize the exponents.

• Our equation is now: 3x − 5 = 34 { displaystyle 3 ^ {x-5} = 3 ^ {4}}

. La base (3) étant identique, ignorez-la et inscrivez l'égalité des exposants: x−5=4{displaystyle x-5=4}

#### Step 4. Solve the equation

First, isolate the unknown on one side. Any operation made on one member of an equation must be done identically on the other member.

• Isolate the unknown by adding 5 to each side of the equation:

x − 5 = 4 { displaystyle x-5 = 4}

x−5+5=4+5{displaystyle x-5+5=4+5}

x=9{displaystyle x=9}

(solution).

#### Step 5. Check your calculations

The verification consists in taking the result found and making the numerical application in the starting equation. After calculations, equality must be verified. If it was not, double check, and if you still come across an inequality, go back to your calculations.

• You found that: x = 9 { displaystyle x = 9}

. Dans l'équation de départ, remplacez x{displaystyle x}

par 9 et faites les calculs:

3x−5=81{displaystyle 3^{x-5}=81}

39−5=81{displaystyle 3^{9-5}=81}

34=81{displaystyle 3^{4}=81}

81=81{displaystyle 81=81}

### Méthode 3 sur 3: Trouver une inconnue en exposant (bases différentes)

#### Step 1. Isolate the power in the left limb

If this is not already the case in your starting equation, isolate the power on the left and the constants on the right, keeping in mind that any operation on a member of an equation must be done identically on the other member.

• Take as an example the equation 43 + x − 8 = 17 { displaystyle 4 ^ {3 + x} -8 = 17}

. Isolez 43+x{displaystyle 4^{3+x}}

dans le membre de gauche en ajoutant de chaque côté 8:

43+x−8=17{displaystyle 4^{3+x}-8=17}

43+x−8+8=17+8{displaystyle 4^{3+x}-8+8=17+8}

43+x=25{displaystyle 4^{3+x}=25}

#### Step 2. Edit the equation

17 cannot be expressed as a power of 4, as was done in the previous method. We must therefore use the decimal logarithm, the latter being a reciprocal function of f (x) = 10x. Take the log of each member. In doing so, the equality remains perfect, since the same operation was performed on both limbs. For calculations, use a scientific calculator.

• Starting from 43 + x = 25 { displaystyle 4 ^ {3 + x} = 25}

, modifiez l'équation, mais elle restera inchangée, en mettant à égalité leurs logs: log(43+x)=log(25){displaystyle {text{log}}(4^{3+x})={text{log}}(25)}

#### Step 3. Take the exponent out of the power

There is a log rule that says: log (xy) = ylog (x) { displaystyle { text {log}} (x ^ {y}) = y { text {log}} (x)}

. Utilisez cette règle pour donner à votre équation un air plus familier. Pour l'instant, il n'est pas question de calculer les logs.

• C'est ainsi que log(43+x)=log(25){displaystyle {text{log}}(4^{3+x})={text{log}}(25)}
• , conformément à la règle vue précédemment, peut s'écrire: (3+x)log(4)=log(25){displaystyle (3+x){text{log}}(4)={text{log}}(25)}

#### Step 4. Isolate the unknown on the left

As has been seen previously, we must now isolate the unknown on the left and the constants on the right, the latter being particular insofar as they are logs, which are nothing other than numerical values. Divide each member by the log that affects the unknown. If there were any constants, you would group them all to the right and calculate them.

• To isolate x { displaystyle x}

dans (3+x)log(4)=log(25){displaystyle (3+x){text{log}}(4)={text{log}}(25)}

, divisez des deux côtés par log(4){displaystyle {text{log}}(4)}

, puis soustrayez de même 3, ce qui donne les calculs suivants:

(3+x)log(4)=log(25){displaystyle (3+x){text{log}}(4)={text{log}}(25)}

(3+x)log(4)log(4)=log(25)log(4){displaystyle (3+x){frac {{text{log}}(4)}{{text{log}}(4)}}={frac {{text{log}}(25)}{{text{log}}(4)}}}

3+x=log(25)log(4){displaystyle 3+x={frac {{text{log}}(25)}{{text{log}}(4)}}}

3+x−3=log(25)log(4)−3{displaystyle 3+x-3={frac {{text{log}}(25)}{{text{log}}(4)}}-3}

x=log(25)log(4)−3{displaystyle x={frac {{text{log}}(25)}{{text{log}}(4)}}-3}

#### Step 5. Calculate the value of each log

This is where the scientific calculator comes in, because it is a complicated calculation with a table. Locate the log key. Enter the value you want to log first, then press that log key. Calculate all the logs and rewrite the equation, rightly so, with those results. Often the logs need to be rounded.

• To calculate log (25) { displaystyle { text {log}} (25)}

, entrez 25 et appuyez sur log: le résultat est 1, 3979. Faites de même avec log(4){displaystyle {text{log}}(4)}

, tapez 4, appuyez sur log et vous obtenez environ 0, 602. Votre équation devient maintenant élémentaire: x=1, 39790, 602−3{displaystyle x={frac {1, 3979}{0, 602}}-3}

#### Step 6. Do the math

The final result is not very far now, it will necessarily be approximated, because we have rounded the logarithmic values, as well as the result of the fraction. Start by calculating the fraction, then the other operations. If you no longer remember the order of operations, that is, the priorities in performing arithmetic operations, read this article first.

• So we have: x = 1, 39790, 602−3 { displaystyle x = { frac {1, 3979} {0, 602}} - 3}

. calculez la fraction, puis faites la soustraction, ce qui donne:

x=1, 39790, 602−3{displaystyle x={frac {1, 3979}{0, 602}}-3}

x=2, 322−3{displaystyle x=2, 322-3}

x=−0, 678{displaystyle x=-0, 678}

(solution).