Question 378815
{{{5^x = 125^2^(x-1)}}}
Is this really the equation? Or is it:
{{{5^x = 125^(2(x-1))}}}
or
{{{5^x = (125^2)^(x-1)}}}
The first equation is extremely difficult. The other two (which are equivalent to each other) are more like the problems given in Math classes. I am going to assume that the last one is correct.<br>
Usually, when the variable is in an exponent, logarithms are used to solve the equation. The exception is when both sides of the equation can be expressed as powers of the same number.<br>
Since {{{125 = 5^3}}} this equation is one of those that can be solved without logarithms. Substituting {{{5^3}}} for 125 we get:
{{{5^x = ((5^3)^2)^(x-1)}}}
The rule for exponents when raising a power to a power is to multiply the exponents. Using this rule (twice) to simplify the right side:
{{{5^x = (5^6)^(x-1)}}}
{{{5^x = 5^(6x-6)}}}
Both sides of the equation are now powers of 5. In order for these powers of 5 to be equal, the exponents must be equal. So:
x = 6x-6
This is an easy equation to solve. Subtracting 6x from each side:
-5x = -6
Dividing both sides by -5 we get:
{{{x = 6/5}}}<br>
{{{3^(2x+1) = 7^(x-1)}}}
In this equation we cannot express both sides as powers of the same number. So we will need to use logarithms. Logarithms of any base can be used. However, there are two things to consider when selecting a base:<ul><li>Using a base for the logarithm that matches the base of one of the exponents will lead to a simpler expression for an answer.</li><li>Using a base for the logarithm that your calculator "knows", like base 10 or base e (aka ln), will make it easier to find a decimal approximation of the answer.</li></ul>
So using base 3 or base 7 logarithms will make our answer simpler. But using base 10 or base e logarithms will make finding a decimal approximation easier. I am going to choose base 7:
{{{log(7, (3^(2x+1))) = log(7, (7^(x-1)))}}}
Now we use a property of logarithms, {{{log(a, (p^q)) = q*log(a, (p))}}}, to move the exponent of the argument out in front of the logarithm. It is this property which is the very reason we use logarithms on these equations. It allows us to move the exponent, where the variable is, to a place where we can use "regular" Algebra to solve it. Using this property on both sides we get:
{{{(2x+1)*log(7, (3)) = (x-1)*log(7, (7))}}}
By definition, {{{log(7, (7)) = 1}}} so this becomes:
{{{(2x+1)*log(7, (3)) = x-1}}}
On the left side we will use the Distributive Property to multiply:
{{{2x*log(7, (3)) + log(7, (3)) = x-1}}}
Gathering the x terms on one side and the other terms on the other side (by subtracting x and {{{log(7, (3))}}} from each side) we get:
{{{2x*log(7, (3)) - x = -1 - log(7, (3))}}}
Factoring out x we get:
{{{x*(2*log(7, (3)) - 1) = -1 - log(7, (3))}}}
And dividing both sides by {{{(2*log(7, (3)) - 1)}}}:
{{{x = (-1 - log(7, (3)))/(2*log(7, (3)) - 1)}}}
This is an exact expression for the solution to the equation. If you want a decimal approximation it is not too late. We can use the base conversion formula, {{{log(a, (p)) = log(b, (p))/log(b, (a))}}}, to convert the base 7 logarithms into either base 10 or base e logarithms.<br>
For the most part I will leave
{{{e^(4x) = 9}}}
for you to solve. One hint: There is a clear choice for the base of the logarithm to use. Base e logarithms will make the exact answer simpler (since it matches the base of the only exponential term) <b>and</b> it is a base your calculator "knows" so a decimal approximation of the answer will be easy.