Question 813935
log(2x+9)=1+log(x-9)<br>
A general procedure for solving these kinds of equations:<ol><li>Use algebra and/or properties of logarithms to transform the equation into one of the following forms:<ul><li>log(expression) = other-expression</li><li>log(expression) = log(other-expression) (Note: The bases of the two logs must match.)</li></ul></li><li>Eliminate the logarithms:<ul><li>If the equation is in the first form, "log(expression) = other-expression", rewrite the equation in exponential form.</li><li>If the equation is in the second form, "log(expression) = log(other-expression)", set the arguments equal.</li></ul></li><li>Now that the logs are gone, solve the equation (using techniques which are appropriate for the type of equation it is).</li><li>Check your solution. This is <b>not</b> optional! A check must be made to see if the bases and arguments of all logs are valid. Any "solution" which make any base or an argument invalid must be rejected! (Note: Valid bases are positive but not 1 and valid arguments are positive.)</li></ol>Let's try this on your equation. First we decide which form we think will be easiest to achieve. With the "non-log" term of 1 (on the right side), it would seem that the second, "all-log" form will be harder to reach. So we will aim for the first form.<br>
Stage 1: Transform
To reach this form, all we need to do is find a way to combine all the logs into a single logarithm. We will getting them both on the same side of the equation. Subtracting log(x-9) from each side:
log(2x+9)-log(x-9)=1
Now we can use the {{{log(a, (p)) + log(a, (q)) = log(a, (p*q))}}} property to combine them:
{{{log(((2x+9)/(x-9)))=1}}}
And we have reached the first form.<br>
Stage 2: Eliminate the logs.
With the first form we just rewrite the equation in exponential form. In general {{{log(a, (p)) = n}}} is equivalent to {{{p = a^n}}}. Using this pattern, and the fact that the base of "log" is 10, we get:
{{{(2x+9)/(x-9)=10^1}}}
which simplifies to:
{{{(2x+9)/(x-9)=10}}}<br>
Stage 3: Solve
We'll start by eliminating the fraction (by multiplying each side by (x-9):
{{{2x+9=(x-9)10}}}
which simplifies to:'
{{{2x+9=10x-90}}}
Subtracting 2x:
{{{9=8x-90}}}
Adding 90:
{{{99=8x}}}
Divide by 8:
{{{99/8=x}}}
Stage 4: Check
Use the original equation to check:
log(2x+9)=1+log(x-9)
Checking x = 99/8:
{{{log((2(99/8)+9))=1+log(((77/8)-9))}}}
Simplifying:
{{{log((2(99/8+72/8)))=1+log((77/8-72/8))}}}
{{{log((2(171/8)))=1+log((5/8))}}}
At this point we can see that both arguments are or are going to be positive (i.e. valid). And the bases are valid so this solution checks out! x = 77/8.<br>
I can't help you with the second problem because:<ul><li>You didn't include the instructions. What are you/we supposed to do with this?</li><li>I can't tell if the equation is:
{{{log((6))*sqrt(x) - (log((4)))^2}}}
or
{{{log((6))*sqrt(x) - log((4^2))}}}
or
{{{log(6, (sqrt(x) - (log(4))^2))}}}
etc.
Please<ul><li>Use parentheses generously to group things like function arguments, exponents, numerators and denominators together so that the meaning of the expression cannot be confused. </li><li>If posting logarithms with bases other than 10 ("log") or e ("ln"), either use some English (like "base 6 log of the square root of (x)" or teach yourself how to use algebra.com's formula syntax. Clink on the "Show source" link above to see what I typed to get:
{{{log(5, (x+3))}}}
to display like it does.</li></ul></li></ul>