Question 814488
I'm assuming that the equation is:
{{{log(7, (x))+5 = 12}}}
Tutor don't like having to guess what the problem is. You'll get a faster response if you express the problem clearly.<br>
One way to make the problem clearer is to use parentheses generously. Use them to group things like function arguments, exponents, numerators, denominators, radicands, etc.<br>
Expressing logarithms clearly can be especially difficult. Try either of the following:<ul><li>Use a mix of English and math symbols. For example: One way to express the equation above would be: "(base 7 log of x) + 5 = 12", or ...</li><li>Teach yourself algebra.com's syntax for formulas. Click on the "Show source" link above to see what I typed to get algebra.com to display the equation above. From that you can possibly learn enough to enter logarithms clearly.</li></ul>If I am wrong about what the equation is, then please re-post your question. If I'm correct then keep reading...<br>
To solve this we first isolate the logarithm. Subtracting 5 we get:
{{{log(7, (x)) = 7}}}
Then rewrite it in exponential form. In general {{{log(a, (p)) = q}}} is equivalent to {{{p = a^q}}}. Using this pattern with our equation we get:
{{{x = 7^7}}}
x = 823543<br>
When solving these equation a check must be made. It is <i>not</i> optional! A solution to these equations must make all bases and arguments of all logs valid. (Valid bases are any positive number except 1 and valid arguments are any positive number.) If a "solution" makes a base or an argument invalid then it is not actually a solution and it must be rejected.<br>
Use the original equation to check:
{{{log(7, (x))+5 = 12}}}
Checking x = 823543:
{{{log(7, (823543))+5 = 12}}}
We can already see that the base, 7, and the argument, 823543, are both valid. So this solution checks out!