To solve exponential equations where the variable is in the exponent, express all terms with the same base, apply the laws of indices (a^m ÷ a^n = a^(m-n)), then take logarithms of both sides and use the power rule of logarithms (log(a^m) = m·log(a)) to isolate the variable. For the equation 8^x ÷ 64 = 6, the solution is x = 7/3 + (1/3)log₂(3).
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Solve for x in this nice exponential equation | Can you solve this? | Math Olympiad Mathematics
Added:In this video, we want to solve for X.
Given 8 raised to power X divided by 64 is equal to 6.
We're given 8 raised to power X divided by 64 is equal to 6.
6 here we can write as 2 * 3.
And we can write the left-hand side as 8 raised to power X divided by 2 raised to power 6 is equal to 2 * 3.
Then I will divide both sides by 2.
Which is something as multiplying this side by 1 over 2.
So we have 8 raised to power X divided by this times this will then be 2 raised to power 7.
Then equal to this here takes care of this.
So we have 3 here.
8 is 2 raised to power 3.
Then raised to power X divided by 2 raised to power 7.
Then equal to 3.
By law of indices, the numerator here will give us 2 raised to power 3 X.
Then divided by 2 raised to power 7.
Then equal to 3.
We're applying the law of indices.
Given A raised to the M divided by a raised to the power n, this will give us a raised to the power m minus n.
So, this here becomes 2 raised to the power 3x - 7 is equal to 3.
Since this is an exponential equation, let us take the logarithm of both sides.
So, log 2 raised to the power 3x - 7 is equal to log 3.
The left-hand side expression is of the form log a raised to the power m, and by law of logarithm, this will give us m * log a which would then imply 3x - 7 * log 2 is equal to log 3.
Let's divide both sides by log 2.
Therefore, this here will cancel this.
Then we have 3x - 7 is equal to log 3 divided by log 2.
Log 3 divided by log 2 here is of the form log m divided by log n.
By law of logarithm, this will give us log m base n.
Then 3x - 7 is equal to log 3 base 2.
Then 3 x is equal to 7 plus log 3 base 2.
Let's divide both sides by 3.
That's each of these as well divided by 3.
So, 3 cancels 3 here, leaving us with x is equal to 7 over 3 plus 1 over 3 log 3 base 2.
This will be our final answer to this problem.
Next thing we'll do will be to verify that this is correct by substituting this value of x into the given problem 8 raised to power x divided by 64 to give us 6.
So, in place of x here, we'll put this, which would then imply 8 raised to power x is 7 over 3 plus 1 over 3 log 3 base 2 divided by 64 to give us 6.
Let's separate this power as well.
Applying this property of exponent p raised to power m plus k will give us p raised to power m times p raised to power k.
So, this becomes 8 raised to power 7 over 3 times 8 raised to power 1 over 3 log 3 base 2 divided by 64 to give us six.
This will be cube root of eight raised to power seven times cube root of eight raised to power log three base two all divided by 64 to give us six.
Cube root of eight is two, so this is going to be two raised to power seven times two raised to power log three base two divided by 64 to give us six.
Two raised to power seven is 128.
Then times two raised to power log three base two then divided by 64 to give us six.
64 here is one. 64 in 128 is two.
So, we are left with two times two raised to power log three base two to give us six.
Next, we'll look at two raised to power log three base two, which is of the form K raised to power log M base K and by law of logarithm, this will give us M so that this expression here will give us three.
Then we have two here times three to give us six.
Two times three is definitely six, so we have 6 on the left is equal to 6 on the right.
And since the left-hand side balances the right-hand side, it confirms that the value we got for X, which is 7 over 3 + 1 over 3 log 3 base 2, is perfectly correct.
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