Question 1163250
Pls help me. Given that log2^x+log3^81=1 find x?
<pre>{{{matrix(1,3, log (2, (x)) + log (3, (81)), "=", 1)}}}

Before we go any further, you need to know the value of {{{log (3, (81)))}}}, which we will name c. This gives us: {{{matrix(1,3, c, "=", log (3, (81)))}}}
{{{matrix(1,3, 3^c, "=", 81))}}} ----- Converting to EXPONENTIAL form, based on the RULE: {{{matrix(1,7, c, "=", log (b, (a)), "=====>", b^c, "=", a)}}}
{{{matrix(1,3, 3^c, "=", 3^4)}}}
c = 4 ----- BASES are equal and so are the exponents


Continuing, {{{matrix(1,3, log (2, (x)) + log (3, (81)), "=", 1)}}} now becomes: {{{matrix(1,3, log (2, (x)) + 4, "=", 1)}}}
{{{matrix(2,3, log (2, (x)), "=", 1 - 4,
log (2, (x)), "=", - 3)}}}
{{{matrix(1,3, x, "=", 2^(- 3))}}} ----- Converting to EXPONENTIAL form
{{{highlight_green(matrix(1,5, x, "=", (1/2)^3, "=", highlight(1/8)))}}}</pre>