# Logarithms where you are solving for X

• chemistudent
In summary: So for example, if you have an equation like y=mx+b, then there is also an equation y=mx+b*e which has the same solution (y=mx+b).
chemistudent

## Homework Statement

Put in Exponential form and solve for X. SHOW YOUR STEPS!

log2(x-5)=5

## Homework Equations

I am not sure what this is asking for. There are no other equations.

## The Attempt at a Solution

25=X-5

25+5=X-5+5

32+5=X

X=37

Looks good to me. What's the question?

Dick said:
Looks good to me. What's the question?

How can I check this with a calculator? Because mine only has log10

And if there isn't a way to check this with a calculator, Is there some website or something that i can use? Because when I look I can't find one.

Last edited:
You should be able to take log2(32) in your head. Since 32=2^5. But for future reference, log2(x)=log10(x)/log10(2).

Dick said:
You should be able to take log2(32) in your head. Since 32=2^5. But for future reference, log2(x)=log10(x)/log10(2).

um. I entered it into my calculator and I got an answer of 166.7025077 instead of 5.

I entered (log(37)/log(2))(X-5)

Put x=37. log2(x-5)=log2(37-5)=log2(32)=log10(32)/log10(2)=1.50515/0.30103=5. But again, this isn't a question where you need to rely on a calculator.

Dick said:
Put x=37. log2(x-5)=log2(37-5)=log2(32)=log10(32)/log10(2)=1.50515/0.30103=5. But again, this isn't a question where you need to rely on a calculator.

ok then.

um... one more question

3(X+6)=27

log327=X+6

Aren't these the same thing?

And if they are is this how you would answer this question?

## Homework Statement

Put in Exponential form and solve for X. SHOW YOUR STEPS!

3(X+6)=27

log327=X+6

## Homework Equations

I am not sure what this is asking for. There are no other equations.

## The Attempt at a Solution

(log(27)/log(3))-6=X+6-6

3-6=X

X=-3

Yes, the two equations are equivalent, which means essentially that they are both saying the same thing. Both have the same value of x as their solution.

In the first equation, 3x + 6 = 27, x + 6 represents the exponent on 3 that gives you 27. That means that x + 6 = 3, or equivalently, that x = -3.

In the second equation, log3 27 = x + 6, the left side simplifies to 3, giving you 3 = x + 6, or equivalently, x = -3.

Every time you have an exponential equation, there is an equivalent log equation that has the same solution (as long as the base is positive and not equal to 1). The opposite is also true; namely that every log equation has an exponential equation counterpart.

## What are logarithms and how are they used?

Logarithms are mathematical functions that help us solve exponential equations. They are used to find the unknown value (X) in equations where the base (b) and the result (y) are known. The logarithmic function is written as logb(y) = X.

## How do you solve for X using logarithms?

To solve for X using logarithms, we use the inverse operation of exponentiation. First, we isolate the logarithm on one side of the equation. Then, we raise the base (b) to the power of both sides. This will cancel out the logarithm, leaving us with the value of X.

## What is the difference between natural logarithms and common logarithms?

Natural logarithms use the base e (Euler's number), while common logarithms use the base 10. This means that natural logarithms are useful for solving exponential equations involving natural phenomena such as growth and decay, while common logarithms are useful for calculations involving scales and measurements.

## Can you solve for X in any logarithmic equation?

No, not every logarithmic equation can be solved for X. This is because some equations may have multiple solutions or no solution at all. In these cases, we can use logarithmic properties to manipulate the equation and find a simpler form, but we may not be able to solve for X directly.

## Are there any special rules for solving logarithmic equations?

Yes, there are some special rules that can help us solve logarithmic equations. These include the product rule, quotient rule, and power rule. These rules allow us to simplify expressions with multiple logarithms or exponents, making it easier to solve for X.

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