How do you solve $2^(x+1) = 3^x$ ?

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Answer 1 · 2016-02-26T03:16:57

Convert to logarithmic form.

$log2^(x + 1) = log3^x$

Simplify using the rule $loga^n = nloga$

$(x + 1)log2 = xlog3$

Distribute on the left side. Don't forget you cannot multiply directly a log with a non-log (e.g $2 xx log3!= log6$ but is equal to $2log3$ )

$xlog2 + log2 = xlog3$

Put the x's to one side of the equation.

$log2 = xlog3 - xlog2$

$log2 = x(log3 - log2)$

Simplify the right side of the equation further by using the rule $logm - logn = log(m / n)$

$log2 = x(log(3/2))$

$log2/log(3/2) = x$

You will want to ask your teacher if he/she wants the answer in exact form or rounded off. Just make sure to check.

Practice exercises:

a) $2^(3x) = 5^(x + 1)$

b). $5^(x - 3) = 3^(2x + 1)$

Challenge problem

Find the value of x in $2^(4x - 6) = 5 xx 3^(x + 7)$

How do you solve 2^(x+1) = 3^x2^(x+1) = 3^x?